JEE Type Practice Problems

Hint

Let the centre of circle $C$ be $(h, k)$. Then as this circle touches axis of $x$, its radius $=|k|$

Also, it touches the given circle $x^2+(y-1)^2=1$, centre $(0,1)$ radius 1 , externally
Therefore, the distance between centres = sum of radii
$\begin{array}{lc} \Rightarrow & \sqrt{(h-0)^2+(k-1)^2}=1+|k| \\ \Rightarrow & h^2+k^2-2 k+1=(1+|k|)^2 \\ \Rightarrow & h^2+k^2-2 k+1=1+2|k|+k^2 \\ \Rightarrow & h^2=2 k+2|k| \end{array}$
$\therefore$ Locus of $(h, k)$ is, $x^2=2 y+2|y|$
Now if $y>0$, it becomes $x^2=4 y$
and if $y \leq 0$, it becomes $x=0$
$\therefore$ Combining the two, the required locus is
$\left\{(x, y): x^2=4 y\right\} \cup\{(0, y): y \leq 0\}$

Hint

$\begin{aligned} & s_1=x^2+y^2+2 a x+c y+a=0 \\ & \quad s_2=x^2+y^2-3 a x+d y-1=0 \end{aligned}$
Equation of common chord of circles $s_1$ and $s_2$ is given by
$\begin{array}{ll} s_1-s_2=0 & \\ \Rightarrow & 5 a x+(c-d) y+a+1=0 \end{array}$
Given, that $5 x+b y-a=0$ passes through $P$ and $Q$
$\therefore$ The two equations should represent the same line
$\begin{array}{ll} \Rightarrow & \frac{a}{1}=\frac{c-d}{b}=\frac{a+1}{-a} \\ \Rightarrow \quad & a+1=-a^2 \\ & a^2+a+1=0 \end{array}$
No real value of $a$.

Hint

Equation of circle with centre $(0,3)$ and radius 2 is
$x^2+(y-3)^2=4$
Let centre of the variable circle is $(\alpha, \beta)$
$\because$ It touches $X$-axis.
$\therefore$ It’s equation is $(x-\alpha)^2+(y+\beta)^2=\beta^2$

Circle touch externally
$\begin{array}{ll} \Rightarrow & c_1 c_2=r_1+r_2 \\ \therefore & \sqrt{\alpha^2+(\beta-3)^2}=2+\beta \\ & \alpha^2+(\beta-3)^2=\beta^2+4+4 \beta \\ \Rightarrow & \alpha^2=10(\beta-1 / 2) \end{array}$
$\therefore$ Locus is $x^2=10\left(y-\frac{1}{2}\right)$ which is a parabola.

Hint

Let the centre be $(\alpha, \beta)$
$\because$ It cuts the circle $x^2+y^2=p^2$ orthogonally
$\therefore$ Using $2 g_1 g_2+2 f_1 f_2=c_1+c_2$, we get
$\begin{aligned} & 2(-\alpha) \times 0+2(-\beta) \times 0=c_1-p^2 \\ & \Rightarrow \quad c_1=p^2 \\ & \end{aligned}$
Let equation of circle is
$x^2+y^2-2 \alpha x-2 \beta y+p^2=0$
It passes through
$(a, b) \Rightarrow a^2+b^2-2 \alpha a-2 \beta b+p^2=0$
$\therefore$ Locus of $(\alpha, \beta)$ is
$\therefore \quad 2 a x+2 b y-\left(a^2+b^2+p^2\right)=0$

Hint

Without loss of generally it we can assume the square $A B C D$ with its vertices $A(1,1), B(-1,1), C(-1,-1), D(1,-1)$ $P$ to be the point $(0,1)$ and $Q$ as $(\sqrt{2}, 0)$.

$\begin{aligned} & \text { Then, } \frac{P A^2+P B^2+P C^2+P D^2}{Q A^2+Q B^2+Q C^2+Q D^2} \\ & =\frac{1+1+5+5}{2\left[(\sqrt{2}-1)^2+1\right]+2\left((\sqrt{2}+1)^2+1\right)} \\ & =\frac{12}{16}=0.75 \\ & \end{aligned}$

Hint

Let $C^{\prime}$ be the circle touching circle $C_1$ and $L$, so that $C_1$ and $C^{\prime}$ are on the same side of $L$. Let us draw a line $T$ parallel to $L$ at a distance equal to the radius of circle $C_1$, on opposite side of $L$. Then, the centre of $C^{\prime}$ is equidistant from the centre of $C_1$ and from line $T$.
$\Rightarrow$ locus of centre of $C^{\prime}$ is a parabola.

Hint

Since, $S$ is equidistant from $A$ and line $B D$, it traces a parabola. Clearly, $A C$ is the axis, $A(1,1)$ is the focus and $T_1\left(\frac{1}{2}, \frac{1}{2}\right)$ is the vertex of parabola.
$A T_1=\frac{1}{\sqrt{2}} .$
$T_2 T_3$ = latusrectum of parabola

$\begin{gathered} =4 \times \frac{1}{\sqrt{2}}=2 \sqrt{2} \\ \therefore \quad \text { Area }\left(\Delta T_1 T_2 T_3\right)=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times 2 \sqrt{2}=1 \text { sq units. } \end{gathered}$

Hint

Point of intersection of $3 x-47-7=0$ and $2 x-3 y-5=0$ is $(1,-1)$ which is the centre of the circle and radius $=7$
$\therefore$ Equation is $(x-1)^2+(y+1)^2=49$
$\Rightarrow \quad x^2+y^2-2 x+2 y-47=0$

Hint

Let $M(h, k)$ be the mid-point of chord $A B$ where
$\angle A O B=\frac{2 \pi}{3}$

$\therefore \quad \angle A O M=\frac{\pi}{3}$.
Also, $\quad O M=3 \cos \frac{\pi}{3}=\frac{3}{2}$
$\Rightarrow \quad \sqrt{h^2+k^2}=\frac{3}{2}$
$\Rightarrow \quad h^2+k^2=\frac{9}{4}$
$\therefore$ Locus of $(h, k)$ is $x^2+y^2=\frac{9}{4}$

Hint

Equation of director circle of the given circle $x^2+y^2=169$ is $x^2+y^2=2 \times 169=338$.
We know from every point on director circle, the tangents drawn to given circle are perpendicular to each other. Here, $(17,7)$ lies on director circle.
$\therefore$ The tangent from $(17,7)$ to given circle are mutually perpendicular.

Hint

Equation of circle whose centre is $(h, k)$
$(x-h)^2+(y-k)^2=k^2$

(radius of circle $=k$ because circle is tangent to $x$-axis)
Equation of circle passing through $(-1,+1)$
$\begin{array}{ll} \therefore & (-1-h)^2+(1-k)^2=k^2 \\ \Rightarrow & 1+h^2+2 h+1+k^2-2 k=k^2 \\ \Rightarrow & h^2+2 h-2 k+2=0 D \geq 0 \\ \therefore & (2)^2-4 \times 1 .(-2 K+2) \geq 0 \\ \Rightarrow & 4-4(-2 k+2) \geq 0 \\ \Rightarrow & 1+2 k-2 \geq 0 \Rightarrow k \geq \frac{1}{2} \end{array}$

Hint

Slope of $C D=\frac{1}{\sqrt{3}}$
$\therefore$ Parametric equation of $C D$ is
$\frac{x-\frac{3 \sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=\frac{y-\frac{3}{2}}{\frac{1}{2}}= \pm 1$
$\therefore$ Two possible coordinates of $C$ are
$\left(\frac{\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}, \frac{1}{2}+\frac{3}{2}\right) \text { or }\left(\frac{-\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2},-\frac{1}{2}+\frac{3}{2}\right)$
i.e. $(2 \sqrt{3}, 2)$ or $(\sqrt{3}, 1)$
As $(0,0)$ and $C$ lie on the same side of $P Q$ $\therefore(\sqrt{3}, 1)$ should be the coordinates of $C$.
Remark : Remember $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ lie on the same or opposite side of a line $a x+b y+c=0$ according as $\frac{a x_1+b y_1+c}{a x_2+b y_2+c}>0$ or $<0 . \therefore$ Equation of the circle is $(x-\sqrt{3})^2+(y-1)^2=1$

Hint

$\triangle P Q R$ is an equilateral triangle, the incentre $C$ must coincide with centroid of $\triangle P Q R$ and $D, E, F$ must coincide with the mid points of sides $P Q, Q R$ and $R P$ respectively.
Also, $\quad \angle C P D=30^{\circ} \Rightarrow P D=\sqrt{3}$
Writing the equation of side $P Q$ in symmetric form we get,
$\begin{gathered} \frac{x-\frac{3 \sqrt{3}}{2}}{-\frac{1}{2}}=\frac{y-\frac{3}{2}}{\frac{\sqrt{3}}{2}}=\mp \sqrt{3} \\ \therefore \text { Coordinates of } P=\left(\frac{\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}, \frac{-3}{2}+\frac{3}{2}\right)=(2 \sqrt{3}, 0) \\ \text { and coordinates of } Q=\left(\frac{-\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}, \frac{3}{2}+\frac{3}{2}\right)=(\sqrt{3}, 3) \end{gathered}$
Let coordinates of $R$ be $(\alpha, \beta)$, then using the formula for centriod of $\Delta$ we get
$\frac{\sqrt{3}+2 \sqrt{3}+\alpha}{3}=\sqrt{3} \text { and } \frac{3+0+\beta}{3}=1$
$\Rightarrow \quad \alpha=0$ and $\beta=0$
$\therefore$ Coordinates of $R=(0,0)$
Now, coordinates of $E=$ mid point of $Q R=\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right)$ and coordinates of $F=$ mid-point of $P R=(\sqrt{3}, 0)$

Hint

Equation of side $Q R$ is $y=\sqrt{3} x$ and equation of side $R P$ is
$y=0$

Hint

The given circle is $x^2+y^2+6 x-10 y+30=0$ Centre $(-3,5)$, radius $=2$
$\begin{array}{r} L_1: 2 x+3 y+(p-3)=0 ; \\ L_2: 2 x+3 y+p+3=0 \end{array}$
Clearly, $L_1 \| L_2$
Distance between $L_1$ and $L_2$
$=\left|\frac{p+3-p+3}{\sqrt{2^2+3^2}}\right|=\frac{6}{\sqrt{13}}<2$
$\Rightarrow$ If one line is a chord of the given circle, other line may or may not the diameter of the circle.
Statement $I$ is true and statement $II$ is false.

Hint

The given circle is $x^2+y^2+2 x+4 y-3=0$

Centre $(-1,-2)$
Let $Q(\alpha, \beta)$ be the point diametrically opposite to the point $P(1,0)$,
then, $\frac{1+\alpha}{2}=-1$ and $\frac{0+\beta}{2}=-2$
$\Rightarrow \quad \alpha=-3, \beta=-4 \text {, So, } Q \text { is }(-3,-4)$

Hint

Tangents $P A$ and $P B$ are drawn from the point $P(1,3)$ to circle $x^2+y^2-6 x-4 y-11=0$ with centre $C(3,2)$

Clearly the circumcircle of $\triangle P A B$ will pass through $C$ and as $\angle A=90^{\circ}, P C$ must be a diameter of the circle.
$\therefore$ Equation of required circle is
$\Rightarrow \quad \begin{aligned} (x-1)(x-3)+(y-2) & =0 \\ \Rightarrow \quad x^2+y^2-4 x-10 y+19 & =0 \end{aligned}$

Hint

Let $r$ be the radius of required circle.
Clearly, in $\Delta C_1 C C_2, C_1 C=C_2 C=r+1$ and $P$ is mid-point of $C_1 C_2$
$\begin{array}{ll} \therefore & C P \perp C_1 C_2 \\ \text { Also, } & P M \perp C C_1 \end{array}$
Now, $\triangle P M C_1 \sim \triangle C P C_1$ (by $A A$ similarity)
$\therefore \quad \frac{M C_1}{P C_1}=\frac{P C_1}{C C_1}$

$\Rightarrow \quad \frac{1}{3}=\frac{3}{r+1} \Rightarrow r+1=9 \Rightarrow r=8 \text {. }$

Hint

The given circles are
$\begin{aligned} & S_1 \equiv x^2+y^2+3 x+7 y+2 p-5=0 \dots(i) \\ & S_2 \equiv x^2+y^2+2 x+2 y-p^2=0 \dots(ii) \end{aligned}$
$\therefore$ Equation of common chord $P Q$ is $S_1-S_2=0$
$\Rightarrow \quad L \equiv x+5 y+p^2+2 p-5=0$
$\Rightarrow$ Equation of circle passing through $P$ and $Q$ is $S_1+\lambda L=0$
$\Rightarrow\left(x^2+y^2+3 x+7 y+2 p-5\right)+\lambda\left(x+5 y+p^2+2 p-5\right)=0$
As it passes through $(1,1)$, therefore
$\begin{array}{ll} \Rightarrow & (7+2 p)+\lambda\left(2 p+p^2+1\right)=0 \\ \Rightarrow \quad & \lambda=-\frac{2 p+7}{(p+1)^2} \end{array}$
which does not exist for $p=-1$

Hint

Circle $x^2+y^2-4 x-8 y-5=0$
$\begin{aligned} & \text { Centre }=(2,4) \\ & \text { Radius }=\sqrt{4+16+5}=5 \end{aligned}$
If circle is intersecting line $3 x-4 y=m$, at two distinct points. $\Rightarrow$ length of perpendicular from centre to the line $<$ radius
$\begin{array}{lc} \Rightarrow & \frac{|6-16-m|}{5}<5 \\ \Rightarrow & |10+m|<25 \\ \Rightarrow & -25<m+10<25 \\ \Rightarrow & -35<m<15 \end{array}$

Hint

$\Rightarrow \quad(-1-h)^2+4=h^2 \Rightarrow h=\frac{-5}{2}$
$\therefore \quad$ Centre $\left(\frac{-5}{2}, 2\right)$ and $r=\frac{5}{2}$
Distance of centre from $(-4,0)$ is $\frac{5}{2}$
$\therefore$ It lies on the circle.

Hint

The smaller region of circle is the region given by
$\begin{array}{ll} & x^2+y^2 \leq 6 \dots(i) \\ \text { and } \quad & 2 x-3 y \geq 1 \dots(ii) \end{array}$

We observe that only two points $\left(2, \frac{3}{4}\right)$ and $\left(\frac{1}{4},-\frac{1}{4}\right)$ satisfy both the inequations Eqs. (i) and (ii), we get $\therefore 2$ points in $S$ lie inside the smaller part.

Hint

As centre of one circle is $(0,0)$ and other circle passes through $(0,0)$, therefore
$\begin{array}{ll} \text { Aslo, } \quad & C_1\left(\frac{a}{2}, 0\right) C_2(0,0) \\ & r_1=\frac{|a|}{2},  r_2=c \\ & C_1 C_2=r_1-r_2=\frac{|a|}{2} \\ \Rightarrow \quad & c-\frac{|a|}{2}=\frac{|a|}{2} \Rightarrow c=|a| \end{array}$
If the two circles touch each other, then they must touch each other internally.

Hint

Any point $P$ on line $4 x-5 y=20$ is $\left(\alpha, \frac{4 \alpha-20}{5}\right)$
Equation of chord of contact $A B$ to the circle $x^2+y^2=9$

drawn from point $P\left(\alpha, \frac{4 \alpha-20}{5}\right)$ is
$x \cdot \alpha+y \cdot\left(\frac{4 \alpha-20}{5}\right)=9 \dots(i)$
Also, the equation of chord $A B$ whose mid-point is $(h, k)$ is
$h x+k y=h^2+k^2 \dots(ii)$
$\because$ Eqs. (i) and (ii) represent the same line, therefore
$\frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^2+k^2}{9}$
$\begin{array}{lrl} \Rightarrow & 5 k \alpha & =4 h \alpha-20 h \\ \text { and } & 9 h & =\alpha\left(h^2+k^2\right) \\ \Rightarrow & \alpha & =\frac{20 h}{4 h-5 k} \text { and } \alpha=\frac{9 h}{h^2+k^2} \end{array}$
$\begin{aligned} & \Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^2+k^2} \\ & \Rightarrow \quad 20\left(h^2+k^2\right)=9(4 h-5 k) \\ & \therefore \text { Locus of }(h, k) \text { is } 20\left(x^2+y^2\right)-36 x+45 y=0 \end{aligned}$

Hint

Equation of tangent $P T$ to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$ is $x \sqrt{3}+y=4$
Let the line $L$, perpendicular to tangent $P T$ be
$x-y \sqrt{3}+\lambda=0 \dots(i)$
As it is tangent to the circle $(x-3)^2+y^2=1$
$\therefore$ length of perpendicular from centre of circle to the tangent
$=$ radius of circle.
$\Rightarrow \quad\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1$ or -5
From Eq. (i)
Equation of $L$ can be
$\begin{aligned} & x-\sqrt{3} y=1 \\ & x-\sqrt{3} y=5 \end{aligned}$

Hint

Equation of tangent $P T$ to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$ is $x \sqrt{3}+y=4$
Let the line $L$, perpendicular to tangent $P T$ be
$x-y \sqrt{3}+\lambda=0 \dots(i)$
As it is tangent to the circle $(x-3)^2+y^2=1$
$\therefore$ length of perpendicular from centre of circle to the tangent
$=$ radius of circle.
$\Rightarrow \quad\left|\frac{3+\lambda}{2}\right|=1 \Rightarrow \lambda=-1$ or -5
From the figure it is clear that the intersection point of two direct common tangents lies on $X$-axis.
Aslo
$\Delta P T_1 C_1 \sim \Delta P T_2 C_2$
$\Rightarrow \quad P C_1: P C_2=2: 1$
or $P$ divides $C_1 C_2$ in the ratio $2: 1$ externally
$\therefore$ Coordinates of $P$ are $(6,0)$
Let the equation of tangent through $P$ be
$y=m(x-6)$
As it touches $x^2+y^2=4$
$\begin{aligned} & \therefore \quad\left|\frac{6 m}{\sqrt{m^2+1}}\right|=2 \\ & \Rightarrow \quad 36 m^2=4\left(m^2+1\right) \\ & \Rightarrow \quad m= \pm \frac{1}{2 \sqrt{2}} \\ & \end{aligned}$
$\therefore$ Equations of common tangents are $y= \pm \frac{1}{2 \sqrt{2}}(x-6)$
Also $x=2$ is the common tangent to the two circles.

Hint

Let centre of the circle be $(1,4)$
$[\because$ circle touches $x$-axis at $(1,0)]$

Let the circle passes through the point $B(2,3)$
$\therefore \quad C A=C B \text { (radius) }$
$\begin{array}{lc} \Rightarrow & C A^2=C B^2 \\ \Rightarrow & (1-1)^2+(h-0)^2=(1-2)^2+(h-3)^2 \\ \Rightarrow & h^2=1+h^2+9-6 h \\ \Rightarrow & h=\frac{10}{6}=\frac{5}{3} \end{array}$
Thus, diameter is $2 h=\frac{10}{3}$.

Hint

Since, circle touches $X$-axis at $(3,0)$
$\therefore$ The equation of circle be
$(x-3)^2+(y-0)^2+\lambda y=0$

As it passes through $(1,-2)$
$\therefore$ Put $x=1, y=-2$
$\Rightarrow \quad(1-3)^2+(-2)^2+\lambda(-2)=0 \Rightarrow \lambda=4$
$\therefore$ Equation of circle is $(x-3)^2+y^2-8=0$
Now, from the options $(5,-2)$ satisfies equation of circle.

Hint

There can be two possibilities for the given circle as shown in the figure

$\therefore$ The equations of circle can be
$\begin{array}{ll} & (x-3)^2+(y-4)^2=4^2 \\ \text { or } & (x-3)^2+(y+4)^2=4^2 \\ \text { i.e. } & x^2+y^2-6 x+8 y+9=0 \\ \end{array}$

Hint

Equation of circle $C=(x-1)^2+(y-1)^2=1$
Radius of $T=|y|$
$T$ touches $C$ externally therefore,
Distance between the centres $=$ sum of their radii
$\begin{array}{cc} \Rightarrow & \sqrt{(0-1)^2+(y-1)^2}=1+|y| \\ \Rightarrow & (0-1)^2+(y-1)^2=(1+|y|)^2 \\ \Rightarrow & 1+y^2+1-2 y=1+y^2+2|y| \\ & 2|y|=1-2 y \end{array}$
If $y>0$ then, $2 y=1-2 y \Rightarrow y=\frac{1}{4}$
If $y<0$ then, $-2 y=1-2 y \Rightarrow 0=1$ (not possible)
$\therefore \quad y=\frac{1}{4}$

Hint

Let the equation of circle be
$x^2+y^2+2 g x+2 f y+c=0$
It passes through $(0,1)$
$\therefore \quad 1+2 f+c=0 \dots(i)$
This circle is orthogonal to $(x-1)^2+y^2=16$ i.e.
$\begin{array}{r} x^2+y^2-2 x-15=0 \\ x^2+y^2-1=0 \end{array}$
$\therefore$ We should have
$2 g(-1)+2 f(0)=c-15$
$2 g+c-15=0 \dots(ii)$
Solving Eqs. (i), (ii) and (iii), we get
$c=1, g=7, f=-1$
$\therefore$ Required circle is
$x^2+y^2+14 x-2 y+1=0$
With centre $(-7,1)$ and radius $=7$

Hint

Intersection point of $2 x-3 y+4=0$ and $x-2 y+3=0$ is $(1,2)$

Since, $P$ is the fixed point for given family of lines
So,
$\begin{aligned} & P B=P A \\ & (\alpha-1)^2+(\beta-2)^2=(2-1)^2+(3-2)^2 \\ & (\alpha-1)^2+(\beta-2)^2=1+1=2 \\ & (x-1)^2+(y-2)^2=(\sqrt{2})^2 \end{aligned}$
$(x-a)^2+(y-b)^2=r^2$
Therefore, given locus is a circle with centre $(1,2)$ and radius $\sqrt{2}$.

Hint

$x^2+y^2-4 x-6 y-12=0 \dots(i)$
Centre $C_1=(2,3)$ and Radius, $r_1=5$ units
$x^2+y^2+6 x+18 y+26=0 \dots(ii)$
$\begin{aligned} & \text { Centre, } C_2=(-3,-9) \\ & \text { and radius, } r_2=8 \text { units } \\ & \left|C_1 C_2\right|=\sqrt{(2+3)^2+(3+9)^2}=13 \text { units } \\ & r_1+r_2=5+8=13 \\ & \therefore\left|C_1 C_2\right|=r_1+r_2 \\ & \end{aligned}$
Therefore, there are three common tangents.

Hint

For the given circle, centre $:(4,4)$, radius $=6$
$\begin{aligned} 6+k & =\sqrt{(h-4)^2+(k-4)^2} \\ (h-4)^2 & =20 k+20 \end{aligned}$
$\therefore$ locus of $(h, k)$ is $(x-4)^2=20(y+1)$, which is parabola.

Hint

Centre of $S: O(-3,2)$ and centre of given circle is $A(2,-3)$
$\Rightarrow \quad O A=5 \sqrt{2}$
Also, $\quad A B=5 \quad(\because A B=$ radius of the given circle)
Now, in $\triangle O A B$,
$\begin{aligned} & (O B)^2=(A B)^2+(O A)^2=25+50=75 \\ & \therefore \quad O B=5 \sqrt{3} \\ & \end{aligned}$

Hint

Circle : $x^2+y^2=1$
Equation of tangent at $P(\cos \theta, \sin \theta)$
$x \cos \theta+y \sin \theta=1 \dots(i)$
Equation of normal at $P$
$y=x \tan \theta \dots(ii)$
Equation of tangent at $S$ is $x=1$
$\therefore \quad Q\left(1, \frac{1-\cos \theta}{\sin \theta}\right)=Q\left(1, \tan \frac{\theta}{2}\right)$
$\therefore$ Equation of line through $Q$ and parallel to $R S$ is $y=\tan \frac{\theta}{2}$
$\therefore$ Intersection point $E$ of normal and $y=\tan \frac{\theta}{2}$
$\tan \frac{\theta}{2}=x \tan \theta$
$\begin{array}{cc} \Rightarrow & x=\frac{1-\tan ^2 \frac{\theta}{2}}{2} \\ \therefore & \text { Locus of } E: x=\frac{1-y^2}{2} \text { or } y^2=1-2 x \end{array}$
It is satisfied by the points $\left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$ and $\left(\frac{1}{3}, \frac{-1}{\sqrt{3}}\right)$.

Hint

(2) Equation of circle can be written as
$(x+1)^2+(y+2)^2=p+5 \dots(i)$

Case I. For $p=0$, circle passes through origin and cuts $x$-axis and $y$-axis at $(-2,0)$ and $(0,-4)$ respectively.
Case II. If circle touch X-axis, then
$(1)^2=-p \Rightarrow p=-1$
From Eq. (i), we get
$(x+1)^2+(y+2)^2=2^2$
Cut off $Y$-axis at (put $x=0$ )
$\begin{aligned} &(y+2)^2=3 \\ & \Rightarrow y=-2 \pm \sqrt{3} \\ & \text { or }(0,-2 \pm \sqrt{3}) \end{aligned}$
Case III. If circle touch $Y$-axis, then
$(2)^2=-p$
$\Rightarrow \quad p=-4$
From Eq. (i), we get
$(x+1)^2+(y+2)^2=1$
Cut off $X$-axis at (put $y=0$ )
$(x+1)^2=-3 \text { (impossible) }$

Hint

Equation of tangent at $(1,7)$ to $y=x^2+6$
$\Rightarrow \quad \frac{1}{2}(y+7)=x \cdot 1+6$
$\Rightarrow \quad y=2 x+5 \dots(i)$
This tangent also touches the circle.
$x^2+y^2+16 x+12 y+c=0 \dots(ii)$
Now, solving Eqs. (i) and (ii), we get
$\Rightarrow \quad \begin{aligned} x^2+(2 x+5)^2+16 x+12(2 x+5)+c & =0 \\ \Rightarrow \quad 5 x^2+60 x+85+c & =0 \end{aligned}$
Since, roots are equal, so
$\begin{aligned} & B^2-4 A C=0 \\ & \Rightarrow \quad(60)^2-4 \times 5 \times(85+c)=0 \\ & \Rightarrow \quad 85+c=180 \\ & \Rightarrow \quad 5 x^2+60 x+180=0 \\ & \Rightarrow \quad x=-\frac{60}{10}=-6 \quad \Rightarrow \quad y=-7 \\ & \end{aligned}$
Hence, point of contact is $(-6,-7)$.

Hint

$P \equiv(1,0)$, let $Q \equiv(h, k)$
such that $\quad k^2=8 h \dots(i)$
Let $(\alpha, \beta)$ be the mid-point of $P Q$.
$\begin{array}{ll} \therefore & \alpha=\frac{h+1}{2}, \beta=\frac{k+0}{2} \\ \Rightarrow & h=2 \alpha-1, k=2 \beta \end{array}$
From Eq. (i), we get
$\begin{aligned} (2 \beta)^2 & =8(2 \alpha-1) \\ \Rightarrow \quad \beta^2 & =4 \alpha-2 \\ \Rightarrow \quad \beta^2-4 \alpha+2 & =0 \end{aligned}$
$\therefore$ Required locus is $y^2-4 x+2=0$.

Hint

Coordinates of $S$ are $\left(2 \sqrt{2} \cos 45^{\circ}, 2 \sqrt{2} \sin 45^{\circ}\right)$ i.e. $(2,2)$.

$\begin{aligned} & \therefore \quad S P=P M \\ & \Rightarrow \quad(S P)^2=(P M)^2 \\ & \Rightarrow \quad(x-2)^2+(y-2)^2=\left[\frac{(x+y)}{\sqrt{2}}\right]^2 \\ & \Rightarrow \quad 2\left(x^2+y^2-4 x-4 y+8\right)=x^2+y^2+2 x y \\ & \Rightarrow x^2+y^2-2 x y-8 x-8 y+16=0 \\ & \therefore \quad(x-y)^2=8(x+y-2) \\ & \end{aligned}$

Hint

Equation of tangent to $y=x^2$ is
$y=m x-\frac{1}{4} m^2 \dots(i)$
Equation of tangent to $(x-2)^2=-y$ is
$y=m(x-2)+\frac{1}{4} m^2 \dots(ii)$
$\because$ Eqs. (i) and (ii) are identical.
$\Rightarrow \quad m=0 \text { or } 4$
$\therefore$ Common tangents are $y=0$ and $y=4 x-4=4(x-1)$.

Hint

Given parabola is
$y=\frac{a^3 x^2}{3}+\frac{a^2 x}{2}-2 a \dots(i)$
For vertex $\frac{d y}{d x}=0 \Rightarrow x=-\frac{3}{4 a}$
Substitute $x=-\frac{3}{4 a}$ in Eq. (i), we get
$y=-\frac{35 a}{16}$
$\therefore$ Coordinates of vertex are $\left(-\frac{3}{4 a},-\frac{35 a}{16}\right)$.
For locus let $x=-\frac{3}{4 a}$ and $y=-\frac{35 a}{16}$.
$\therefore \quad x y=\frac{105}{64}$, which is the required locus.

Hint

$y=x^2-5 x+6$
$\therefore$ Equation of tangent at $(2,0)$ is
$\begin{aligned} \frac{y+0}{2} & =x \cdot 2-\frac{5}{2}(x+2)+6 \\ \Rightarrow \quad y & =-x+2 \dots(i) \end{aligned}$
and equation of tangent at $(3,0)$ is
$\Rightarrow \quad y=x-3$
$\because$ Eqs. (i) and (ii) are perpendicular.
$\therefore$ Angle between tangents is $\pi / 2$.

Hint

(i) Coordinates of $P$ and $Q$ are $(1,2 \sqrt{2})$ and $(1,-2 \sqrt{2})$.

Area of $\triangle P Q R=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 8=16 \sqrt{2}$
Area of $\triangle P Q S=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 2=4 \sqrt{2}$
$\therefore$ Ratio of area of $\triangle P Q S$ and $\triangle P Q R$ is $1: 4$.
(ii) Equation of circumcircle of $\triangle P R S$ is
$(x+1)(x-9)+y^2+\lambda y=0$
It will pass through $(1,2 \sqrt{2})$, then
$-16+8+\lambda 2 \sqrt{2}=0 \Rightarrow \lambda=\frac{8}{2 \sqrt{2}}=2 \sqrt{2}$
Equation of circumcircle is
$x^2+y^2-8 x+2 \sqrt{2} y-9=0$
Hence, radius is $3 \sqrt{3}$.
Alternate :
$\begin{aligned} & \text { Let } \angle P S R=\theta \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}} \\ & \Rightarrow \quad P R=6 \sqrt{2}=2 R \cdot \sin \theta \Rightarrow R=3 \sqrt{3} . \end{aligned}$
(iii) Radius of incircle is $r=\frac{\Delta}{ s }$.
$\begin{aligned} \text { As } & \Delta=16 \sqrt{2} \\ & \therefore \quad s=\frac{6 \sqrt{2}+6 \sqrt{2}+4 \sqrt{2}}{2}=8 \sqrt{2} \\ & \therefore \quad r=\frac{16 \sqrt{2}}{8 \sqrt{2}}=2 \\ & \end{aligned}$

Hint

$y=-\frac{x^2}{2}+x+1 \Rightarrow y-\frac{3}{2}=-\frac{1}{2}(x-1)^2$
$\Rightarrow$ It is symmetric about $x=1$.
Hence, both statement are true and Statement II is correct explanation of Statement I.

Hint

Point of intersection of two perpendicular tangents to the parabola lies on directrix of the parabola.
$\therefore$ Equation of directrix is $x+2=0$.
So, point is $(-2,0)$.

Hint

The circle and the parabola touch each other at $x=1$, i.e. at the points $(1,2)$ and $(1,-2)$ as shown in figure.

Hint

Focus of the parabola is $(0,0)$ and $x=2$ is the directrix
And we know that axis of the parabola is perpendicular to the directrix and pass through the focus of the parabola
Thus foot of directrix is $(2,0)$
Also we know that, vertex of the parabola is the mid point of its focus and the foot of directrix
So vertex of parabola is $\left(\frac{0+2}{2}, \frac{0+0}{2}\right)=(1,0)$

Hint

$\begin{aligned} G & \equiv(h, k) \\ \Rightarrow \quad h & =\frac{2 a+a t^2}{3}, k=\frac{2 a t}{3} \end{aligned}$

$\Rightarrow \quad\left(\frac{3 h-2 a}{a}\right)=\frac{9 k^2}{4 a^2}$
$\therefore$ Required parabola is
$\begin{aligned} & \quad \frac{9 y^2}{4 a^2}=\frac{(3 x-2 a)}{a}=\frac{3}{a}\left(x-\frac{2 a}{3}\right) \\ & \Rightarrow \quad y^2=\frac{4 a}{3}\left(x-\frac{2 a}{3}\right) \\ & \therefore \text { Vertex } \equiv\left(\frac{2 a}{3}, 0\right) \text {; Focus } \equiv(a, 0) . \end{aligned}$

Hint

$\text { Slope of } A B=\frac{2 t_2-2 t_1}{\left(t_2^2-t_1^2\right)}=\frac{2}{\left(t_2+t_1\right)} \dots(i)$

$\begin{aligned} & M=\text { Mid-point of } A B=\left(\frac{t_1^2+t_2^2}{2}, t_1+t_2\right) \\ \therefore & r=\left|t_1+t_2\right| \Rightarrow t_1+t_2= \pm r \end{aligned}$
Now, from Eq. (i),
$\text { slope of } A B= \pm \frac{2}{r} \text {. }$

Hint

Given equation of parabola is $y^2=4 x$
We know that the locus of point $P$ from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola.
The standard equation of parabola $(y-k)^2=4 p(x-h)$ has focus $(h+p, k)$ and the directrix is $x = h – p$
From (i), we get
$h = 0 , k = 0 , p =1$
$\Rightarrow$ Directrix of (i) is $x=0-1=-1$
Hence, required locus is $x =- 1$.

Hint

$\begin{aligned} \Delta_1 & =\text { Area of } \triangle P L L^{\prime} \\ & =\frac{1}{2} \times 8 \times\left(2-\frac{1}{2}\right)=6 \text { sq units } \end{aligned}$

Now, equation of $A B$ is $y=2 x+1$, equation of $A C$ is $y=x+2$ and equation of $B C$ is $y=-x-2$
On solving above equations, we get
$A(1,3), B(-1,-1) \text { and } C(-2,0)$
$\begin{array}{ll} \therefore & \Delta_2=\frac{1}{2}\left\|\begin{array}{cc} 1+2 & 3-0 \\ -1+2 & -1-0 \end{array}\right\|=3 \text { sq units } \\ \therefore & \frac{\Delta_1}{\Delta_2}=2 \end{array}$

Hint

Let $A(x, y)=A\left(t^2, 2 t\right)$ be any point on the parabola $y^2=4 x$, then
$x=\frac{t^2}{4} \dots(i)$
and $y=\frac{2 t}{4} \dots(ii)$

From Eqs. (i) and (ii), we get
$x=y^2$

Hint

The equation of normal to
$y^2=4 x \text { is } y=m x-2 m-m^3 \dots(i)$
As it passes through $(9,6)$, then
$6=9 m-2 m-m^3$
$\begin{aligned} & \Rightarrow \quad m^3-7 m+6=0 \\ & \Rightarrow(m-1)(m-2)(m+3)=0 \\ & \Rightarrow \quad m=1,2,-3 \\ & \end{aligned}$
From Eq. (i), equations of normals are
$\begin{aligned} & y=x-3, y=2 x-12, y=-3 x+33 \\ \Rightarrow & y-x+3=0, y-2 x+12=0, y+3 x-33=0 \end{aligned}$

Hint

The shortest distance between $y=x-1$ and $y^2=x$ is along the normal of $y^2=x$.

Let $P\left(t^2, t\right)$ be any point on $y^2=x$.
$\therefore \quad$ Tangent at $P$ is $y=\frac{x}{2 t}+\frac{t}{2}$.
$\therefore \quad$ Slope of tangent $=\frac{1}{2 t}$
and tangent at $P$ is parallel to $y-x=1$
$\therefore \quad \frac{1}{2 t}=1 \Rightarrow t=\frac{1}{2} \Rightarrow P\left(\frac{1}{4}, \frac{1}{2}\right)$
Hence, shortest distance $=P Q=\frac{\left|\frac{1}{2}-\frac{1}{4}-1\right|}{\sqrt{(1+1)}}=\frac{3}{4 \sqrt{2}}=\frac{3 \sqrt{2}}{8}$

Hint

We observe that both parabola $y^2=8 x$ and circle $x^2+y^2-2 x-4 y=0$ pass through origin say $P(0,0)$.

Let $Q$ be the point $\left(2 t^2, 4 t\right)$, then it will satisfy the equation of circle.
$\begin{array}{lc} \therefore & \left(2 t^2\right)^2+(4 t)^2-2\left(2 t^2\right)-4(4 t)=0 \\ \Rightarrow & 4 t^4+12 t^2-16 t=0 \\ \Rightarrow & t(t-1)\left(t^2+t-4\right)=0 \Rightarrow t=0 \text { or } 1 \end{array}$
For $t=0$, we get point $P$, therefore $t=1$ gives point $Q$ as $(2,4)$. Here, $P(0,0)$ and $Q(2,4)$ are end points of diameter of the given circle and focus of the parabola is the point $S(2,0)$.
$\therefore \quad \angle P S Q=90^{\circ}$
Hence, area of $\triangle P Q S=\frac{1}{2} \times 2 \times 4=4$ sq units.

Hint

$\because P Q$ is the focal chord of $y^2=4 a x$.
$\therefore$ Coordinates of $P$ and $Q$ are $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

Tangents at $P$ and $Q$ are
$t y=x+a t^2 \text { and } t y=x t^2+a$
which intersect each other at $R\left(-a, a\left(t-\frac{1}{t}\right)\right)$.
As $R$ lies on the line $y=2 x+a, a>0$
$\begin{aligned} & \therefore \quad a\left(t-\frac{1}{t}\right)=-2 a+a & \Rightarrow \quad t-\frac{1}{t}=-1 \end{aligned}$
$\because$ Slope of $O P=\frac{2}{t}$ and slope of $O Q=-2 t$
$\begin{aligned} & \therefore \quad \tan \theta=\left|\frac{\frac{2}{t}+2 t}{1-4}\right|=\frac{2}{3}\left|t+\frac{1}{t}\right| \\ & =\frac{2}{3} \sqrt{\left(t-\frac{1}{t}\right)^2+4}=\frac{2}{3} \sqrt{5} \quad\left[\because t-\frac{1}{t}=-1\right] \\ & \because \quad \theta>90^{\circ} & \therefore \quad \tan \theta=-\frac{2}{3} \sqrt{5}  & \end{aligned}$
$\because$ Slope of $O P=\frac{2}{t}$ and slope of $O Q=-2 t$
$\begin{aligned} & \therefore \quad \tan \theta=\left|\frac{\frac{2}{t}+2 t}{1-4}\right|=\frac{2}{3}\left|t+\frac{1}{t}\right| \\ & =\frac{2}{3} \sqrt{\left(t-\frac{1}{t}\right)^2+4}=\frac{2}{3} \sqrt{5} \quad\left[\because t-\frac{1}{t}=-1\right] \\ & \because \quad \theta>90^{\circ} & \therefore \quad \tan \theta=-\frac{2}{3} \sqrt{5} \\ & \end{aligned}$

Hint

$\because P Q$ is the focal chord of $y^2=4 a x$.
$\therefore$ Coordinates of $P$ and $Q$ are $\left(a t^2, 2 a t\right)$ and $\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

Tangents at $P$ and $Q$ are
$t y=x+a t^2 \text { and } t y=x t^2+a$
which intersect each other at $R\left(-a, a\left(t-\frac{1}{t}\right)\right)$.
As $R$ lies on the line $y=2 x+a, a>0$
$\begin{aligned} & \therefore \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad t-\frac{1}{t}=-1 \\ & \end{aligned}$
$\begin{aligned} &\begin{aligned} P Q & =a\left(t+\frac{1}{t}\right)^2=a\left\{\left(t-\frac{1}{t}\right)^2+4\right\} \\ & =a(1+4)=5 a \end{aligned}\\ &\left[\because t-\frac{1}{t}=-1\right] \end{aligned}$

Hint

Equation of tangent of $y^2=4 x$ in terms of slope is
$y=m x+\frac{1}{m} \dots(i)$
$\because$ Line Eq. (i) touches $x^2=-32 y$
$\begin{aligned} \Rightarrow & x^2 =-32\left(m x+\frac{1}{m}\right) \\ \Rightarrow & x^2+32 m x+\frac{32}{m} =0 \dots(ii) \end{aligned}$
For touching roots of Eq. (ii) are equal.
$\begin{aligned} & \therefore \quad D=0 \\ & \Rightarrow \quad(32 m)^2=4 \cdot 1 \cdot\left(\frac{32}{m}\right) \\ & \Rightarrow \quad m^3=\frac{1}{8} \\ & \therefore \quad m=1 / 2 \\ & \end{aligned}$

Hint

Let the tangent to, $y^2=8 x$ be $y=m x+\frac{2}{m}$.

If it is common tangent to parabola and circle, then $y=m x+\frac{2}{m}$ is a tangent to $x^2+y^2=2$.

$\begin{aligned} & & \frac{\frac{2}{m}}{\sqrt{\left(1+m^2\right)}} & =\sqrt{2} \\ \Rightarrow & & \frac{4}{m^2\left(1+m^2\right)} & =2 \\ \Rightarrow & & m^4+m^2-2 & =0 \\ \Rightarrow & & \left(m^2+2\right)\left(m^2-1\right) & =0 \\ & \therefore & & m= \pm 1 \end{aligned}$
$\therefore$ Required tangents are $y=x+2$ and $y=-x-2$.
Their common point is $T(-2,0)$.
Chord of contact $P Q$ to circle is
$\begin{aligned} & & x \cdot(-2)+y \cdot 0 & =2 \\ \Rightarrow & & x & =-1 \end{aligned}$
Hence, coordinates of $P$ and $Q$ are $(-1,1)$ and $(-1,-1)$ and chord of contact $R S$ to parabola is
$\begin{aligned} & y \cdot 0=4(x-2) \\ & \Rightarrow \quad x=2 \\ & \end{aligned}$
Hence, coordinates of $R$ and $S$ are $(2,4)$ and $(2,-4)$.
$\therefore$ Area of trapezium $P Q R S=\frac{1}{2}(2+8) \times 3=15$ sq units

Hint

$\because P Q$ is a focal chord, then $Q\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$.
Also,
$Q R \| P K \Rightarrow m_{Q R}=m_{P K}$
$\begin{aligned} & \Rightarrow \quad \frac{2 a r+\frac{2 a}{t}}{a r^2-\frac{a}{t^2}}=\frac{0-2 a t}{2 a-a t^2} \\ & \Rightarrow \quad \frac{2}{r-\frac{1}{t}}=\frac{-2 t}{2-t^2} \\ & {\left[\because r+\frac{1}{t} \neq 0 \text {, otherwise } Q \text { will coincide with } R\right]} \\ & \Rightarrow \quad 2-t^2=-r t+1 \\ & \therefore \quad r=\frac{t^2-1}{t} \\ & \end{aligned}$

Hint

Tangent at $P$ is $\quad t y=x+a t^2 \dots(i)$
Normal at $S$ is $\quad y+s x=2 a s+a s^3 \dots(ii)$
Putting the value of $x$ from Eq. (i) in Eq. (ii), then
$\begin{aligned} & y+s\left(t y-a t^2\right)=2 a s+a s^3 \\ & \Rightarrow \quad y+(s t) y-a(s t) t=2 a s+a s^3 \\ & \Rightarrow \quad y+y-a t=\frac{2 a}{t}+\frac{a}{t^3} \quad[\because s t=1] \\ & \Rightarrow \quad 2 y=a\left(t+\frac{2}{t}+\frac{1}{t^3}\right) \\ & \therefore \quad y=\frac{a\left(t^2+1\right)^2}{2 t^3} \\ & \end{aligned}$

Hint

Let any point $Q$ on $x^2=8 y$ is $\left(4 t, 2 t^2\right)$ and given $P(h, k)$ divides $O Q$ in the ratio $1: 3$ (internally).

Then, $\quad h=\frac{4 t}{4}=t$ and $k=\frac{2 t^2}{4} \Rightarrow 2 k=h^2$
$\therefore$ Required locus of $P$ is $x^2=2 y$.

Hint

End points of latusrectum of
$y^2=4 x \text { are }(1, \pm 2) .$
Equation of normal to $y^2=4 x$ at $(1,2)$ is
$\begin{aligned} y-2 & =-\frac{2}{2}(x-1) \\ \Rightarrow \quad x+y-3 & =0 \end{aligned}$
As it is tangent to circle $(x-3)^2+(y+2)^2=r^2$
$\therefore \quad \frac{|3-2-3|}{\sqrt{(1+1)}}=r \Rightarrow r^2=2$

Hint

Let $\left(t^2, 2 t\right)$ be any point on $y^2=4 x$. Let $(h, k)$ be image of $\left(t^2, 2 t\right)$ with respect to the line $x+y+4=0$, then
$\frac{h-t^2}{1}=\frac{k-2 t}{1}=\frac{-2\left(t^2+2 t+4\right)}{1+1}$
$\begin{aligned} & \Rightarrow \quad h=-(2 t+4) \text { and } k=-\left(t^2+4\right) \\ & \Rightarrow \quad(k+4)=-\left(\frac{h+4}{-2}\right)^2 \\ & \Rightarrow \quad(h+4)^2=-4(k+4) \\ & \end{aligned}$
Locus of $(h, k)$ is $(x+4)^2=-4(y+4)$.
$\therefore \quad \text { Curve } C \text { is }(x+4)^2=-4(y+4)$
Now, intersection of $C$ with $y=-5$, then
$\begin{array}{lc} & (x+4)^2=-4(-5+4)=4 \\ \therefore & x+4= \pm 2 \Rightarrow x=-6,-2 \\ \therefore & A(-6,-5) \text { and } B(-2,-5) \\ \therefore & A B=4 \end{array}$

Hint

Let $P\left(\frac{t_1^2}{2}, t_1\right)$ and $Q\left(\frac{t_2^2}{2}, t_2\right)$ such that $t_1>0$
$[\because P$ lies in first quadrant $]$
$\because$ Circle with $P Q$ as diameter passes through the vertex $O(0,0)$ of the parabola.
$\begin{aligned} & \therefore \quad \angle P O Q=90^{\circ} \\ & \Rightarrow \text { Slope of } O P \times \text { Slope of } O Q=-1 \\ & \Rightarrow \quad \frac{2}{t_1} \times \frac{2}{t_2}=-1 \\ & \Rightarrow \quad t_1 t_2=-4 \quad\left[\because t_2<0\right] \\ & \end{aligned}$

$\begin{aligned} & \text { Now, area of } \triangle O P Q=3 \sqrt{2} \\ & \Rightarrow \quad \frac{1}{2}\left|\begin{array}{cc} \frac{t_1^2}{2} & t_1 \\ \frac{t_2^2}{2} & t_2 \end{array}\right|=3 \sqrt{2} \\ & \Rightarrow \quad \frac{1}{4} t_1 t_2\left(t_1-t_2\right)= \pm 3 \sqrt{2} \\ & \Rightarrow \quad t_1-t_2= \pm 3 \sqrt{2} \left[\because t_1 t_2=-4\right] \end{aligned}$
$\begin{array}{ll} \Rightarrow & t_1+\frac{4}{t_1}= \pm 3 \sqrt{2} \\ \text { or } & t_1+\frac{4}{t_1}=3 \sqrt{2} \left[\because t_1>0\right] \end{array}$
$\begin{array}{lc} \Rightarrow & t_1^2-3 \sqrt{2} t_1+4=0 \\ \therefore & t_1=\frac{3 \sqrt{2} \pm \sqrt{2}}{2}=2 \sqrt{2}, \sqrt{2} . \\ \therefore & \text { Point } P \text { can be }(4,2 \sqrt{2}) \text { or }(1, \sqrt{2}) . \end{array}$

Hint

Let $P\left(2 t^2, 4 t\right)$ and $C(0,-6)$.
$\therefore \quad(C P)^2=4 t^4+(4 t+6)^2=z \text { say }$
$\begin{aligned} & \therefore \quad \frac{d z}{d t}=0 \\ & \Rightarrow \quad 16 t^3+2(4 t+6) \cdot 4=0 \\ & \Rightarrow \quad t^3+2 t+3=0 \\ & \Rightarrow \quad(t+1)\left(t^2-t+3\right)=0 \\ & \end{aligned}$
$\begin{array}{lc} \therefore & t=-1 \\ \Rightarrow & P(2,-4) \end{array}$
Equation of circle is
$\begin{aligned} (x-2)^2+(y+4)^2 & =(2-0)^2+(-4+6)^2 \\ \Rightarrow x^2+y^2-4 x+8 y+12 & =0 \end{aligned}$

Hint

$\because C_1: x^2+y^2=3$ and parabola $x^2=2 y$, then
$y^2+2 y-3=0 \Rightarrow y=1,-3$
$\therefore \quad P(\sqrt{2}, 1)$ $[\because P$ lies in first quadrant $]$
Now, tangent at $P(\sqrt{2}, 1)$ on the circle $C_1$ is
$x \sqrt{2}+y=3$
Let $Q_2$ or $Q_3(0, \lambda)$
$\begin{aligned} & \therefore \quad \frac{|0+\lambda-3|}{\sqrt{(2+1)}}=2 \sqrt{3} \\ & \Rightarrow \quad|\lambda-3|=6 \\ & \therefore \quad \lambda=9 \text { or }-3 \\ & \Rightarrow Q_2(0,-3) \text { and } Q_3(0,9) \text {. } \\ & \end{aligned}$
Alternate (a) $Q_2 Q_3=12$
Alternate (b) $R_2 R_3=$ Length of external common tangent
$\begin{aligned} & =\sqrt{\left(Q_2 Q_3\right)^2-(2 \sqrt{3}+2 \sqrt{3})^2} \\ & =\sqrt{(144-48}=4 \sqrt{6} \end{aligned}$
Alternate (c) Area of $\triangle O R_2 R_3=\frac{1}{2} \times R_2 R_3 \times \frac{|0+0-3|}{\sqrt{(2+1)}}$
$=\frac{1}{2} \times 4 \sqrt{6} \times \frac{3}{\sqrt{3}}=6 \sqrt{2}$
Alternate (d) Area of $\triangle P Q_2 Q_3=\frac{1}{2} \times Q_2 Q_3 \times \sqrt{2}$
$=\frac{1}{2} \times 12 \times \sqrt{2}=6 \sqrt{2}$

Hint

Let $P\left(t^2, 2 t\right), S(2,8)$ and $r=\sqrt{(4+64-64)}=2$
We know that, shortest distance between two curves lies along their common normal. The common normal will pass through centre of circle.
$\therefore$ Slope of $P S=$ Slope of normal to the parabola $y^2=4 x$ at $P\left(t^2, 2 t\right)$
$\Rightarrow \quad \frac{2 t-8}{t^2-2}=-t \text { or } t^3=8 \Rightarrow t=2$
$\therefore P(4,4)$
Alternate (a) $\quad S P=\sqrt{(2-4)^2+(8-4)^2}=2 \sqrt{5}$
Alternate (b) $\quad S Q=r=2$
$\begin{aligned} \therefore \quad \frac{S Q}{Q P} & =\frac{S Q}{S P-S Q}=\frac{2}{2 \sqrt{5}-2} \\ & =\frac{1}{(\sqrt{5}-1)} \times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{\sqrt{5}+1}{4} \\ \Rightarrow \quad S Q: Q P & =(\sqrt{5}+1): 4 \end{aligned}$
Alternate (c) Equation of normal at $P(4,4)$ is
$\begin{array}{rlrl} & & y-4 & =-\frac{4}{2}(x-4) \\ \Rightarrow & & y-4 & =-2 x+8 \\ \Rightarrow & 2 x+y & =12 \end{array}$
$\therefore$ Intercept on $X$-axis is 6 .
Alternate (d) Slope of tangent at $Q=$ Slope of tangent at
$P=\frac{1}{2}$

Hint

Centre of circle

$\begin{aligned} & C \equiv(0,4-r) \\ & \because \quad C M=r \\ & \therefore \quad \frac{|10-(4-r)|}{\sqrt{2}}=r \\ & \Rightarrow \quad 4-r=r \sqrt{2} \\ & \text { or } \\ & r=\frac{4}{\sqrt{2}+1} \\ & =4(\sqrt{2}-1) \\ & \end{aligned}$

Hint

(a) Equation of chord of parabola $y^2=16 x$ whose mid-point $(h, k)$ is
$\begin{aligned} T & =S_1 \\ k y-8(x+h) & =k^2-16 h \\ 8 x-k y & =8 h-k^2 \dots(i) \end{aligned}$
Now comparing Eq. (i) and $2 x+y=p$, then
$\begin{array}{ll} & \frac{8}{2}=\frac{-k}{1}=\frac{8 h-k^2}{p} \\ \Rightarrow & k=-4 \text { and } 4 p=8 h-k^2 \\ \text { or } & k=-4 \text { and } p=2 h-4 \\ \text { Hence, } & p=2, h=3, k=-4 \end{array}$

Hint

Any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$

It meets coordinate axes at $A(a \sec \theta, 0)$ and $B(0, b \operatorname{cosec} \theta)$.
$\begin{aligned} & \therefore \quad \text { Area of } \triangle O A B & =\frac{1}{2} \times a \sec \theta \times b \operatorname{cosec} \theta \\ \Rightarrow & \Delta =\frac{a b}{\sin 2 \theta} \end{aligned}$
For $\Delta$ to be min, $\sin 2 \theta$ should be max and we know max value of $\sin 2 \theta=1$.
$\therefore \quad \Delta_{\max }=a b \text { sq units. }$

Hint

Let the common tangent to circle $x^2+y^2=16$ and ellipse
$\begin{aligned} x^2 / 25+y^2 / 4 & =1 \text { be } \\ y & =m x+\sqrt{25 m^2+4} \dots(i) \end{aligned}$
As it is tangent to circle $x^2+y^2=16$, we should have
$\begin{array}{rlrl} & & \frac{\sqrt{25 m^2+4}}{\sqrt{m^2+1}} & =4 \quad \begin{array}{r} \text { [Using: length of perpendicular } \\ \text { from }(0,0) \text { to }(1)=4] \end{array} \\ \Rightarrow & & 25 m^2+4 & =16 m^2+16 \\ \Rightarrow & 9 m^2 & =12 \\ \Rightarrow & m & =\frac{-2}{\sqrt{3}} \end{array}$
[Leaving + ve sign to consider tangent in I quadrant]
$\therefore$ Equation of common tangent is
$\begin{aligned} y & =-\frac{2}{\sqrt{3}} x+\sqrt{25 \cdot \frac{4}{3}+4} \\ \Rightarrow \quad y & =-\frac{2}{\sqrt{3}} x+4 \sqrt{\frac{7}{3}} \end{aligned}$
This tangent meets the axes at $A(2 \sqrt{7}, 0)$ and $B\left(0,4 \sqrt{\frac{7}{3}}\right)$.
$\therefore$ Length of intercepted portion of tangent between axes
$\begin{aligned} A B & =\sqrt{(2 \sqrt{7})+\left(4 \sqrt{\frac{7}{3}}\right)^3} \\ & =14 / \sqrt{3} \end{aligned}$

Hint

$\begin{aligned} & \because \angle F B F^{\prime}=90^{\circ} \Rightarrow F B^2+F^{\prime} B^2=F F^{\prime 2} \\ & \therefore\left(\sqrt{a^2 e^2+b^2}\right)^2+\left(\sqrt{a^2 e^2+b^2}\right)^2=(2 a e)^2 \end{aligned}$
$\begin{array}{rlrl} \Rightarrow & & 2\left(a^2 e^2+b^2\right) & =4 a^2 e^2 \\ \Rightarrow & e^2 =\frac{b^2}{a^2} \end{array}$

$\begin{array}{ll} \text { Also, } & e^2=1-b^2 / a^2=1-e^2 \\ \Rightarrow & 2 e^2=1, e=\frac{1}{\sqrt{2}} \end{array}$

Hint

$\begin{aligned} & 2 a e=6 \Rightarrow a e=3 ; 2 b=8 \\ & \Rightarrow \quad b=4 \\ & b^2=a^2\left(1-e^2\right) ; 16=a^2-a^2 e^2 \\ & \Rightarrow \quad a^2=16+9=25 \Rightarrow a=5 \\ & \therefore \quad e=\frac{3}{a}=\frac{3}{5} \\ & \end{aligned}$

Hint

Given, ellipse is $x^2+4 y^2=4$
$\begin{array}{ll} \text { or } & \frac{x^2}{2^2}+\frac{y^2}{1}=1 \Rightarrow a=2, b=1 \\ \therefore & e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} \\ \therefore & a e=\sqrt{3} \end{array}$
As per question $P \equiv\left(a e,-b^2 / a\right)=\left(\sqrt{3},-\frac{1}{2}\right)$
$\begin{aligned} & Q \equiv\left(-a e,-b^2 / a\right)=\left(-\sqrt{3},-\frac{1}{2}\right) \\ & \therefore \quad P Q=2 \sqrt{3} \\ & \end{aligned}$
Now, if $P Q$ is the length of latusrectum to be found, then
$P Q=4 a=2 \sqrt{3} \Rightarrow a=\frac{\sqrt{3}}{2}$
Also, as $P Q$ is horizontal, parabola with $P Q$ as latusrectum can be upward parabola (with vertex at $A$ ) or down ward parabola (with vertex at $A^{\prime}$ ).
For upward parabola, $A r=a=\frac{\sqrt{3}}{2}$
$\therefore$ Coordinates of $A=\left(0,-\left(\frac{\sqrt{3}+1}{2}\right)\right)$
So, Equation of upward parabola is given by
$x^2=2 \sqrt{3}\left(y+\frac{\sqrt{3}+1}{2}\right) \text { or } x^2-2 \sqrt{3} y=3+\sqrt{3} \dots(i)$
For downward parabola $A^{\prime} R=a=\frac{\sqrt{3}}{2}$
$\therefore$ Coordinates of $A^{\prime}=\left(0,-\left(\frac{1-\sqrt{3}}{2}\right)\right)$
So, equation of downward parabola is given by
$\begin{aligned} & x^2=-2 \sqrt{3}\left(y+\frac{1-\sqrt{3}}{2}\right) \\ & x^2+2 \sqrt{3} y=3-\sqrt{3} \dots(ii) \end{aligned}$
$\therefore$ Equation of required parabola is given by Eqs. (i) and (ii).

Hint

Perpendicular distance of directrix from focus
$\begin{aligned} & =\frac{a}{e}-a e=4 \\ \Rightarrow a\left(2-\frac{1}{2}\right) & =4 \Rightarrow a=\frac{8}{3} \end{aligned}$

$\therefore \text { Semi major axis }=8 / 3$

Hint

The given ellipse is $x^2+9 y^2=9$ or $\quad \frac{x^2}{3^2}+\frac{y^2}{1^2}=1$

So, the $A(3,0)$ and $B(0,1)$
$\therefore$ Equation of $A B$ is $\frac{x}{3}+\frac{y}{1}=1$ or $x+3 y-3=0 \dots(i)$
Also, auxillary circle of given ellipse is
$x^2+y^2=9 \dots(ii)$
Solving Eqs. (i) and (ii), we get the point $M$ where line $A B$ meets the auxillary circle.
Putting $x=3-3 y$ from Eqs. (i) and (ii)
$\begin{array}{rlrl} \text { we get } & (3-3 y)^2+y^2 =9 \\ \Rightarrow & 9-18 y+9 y^2+y^2 =9 \\ \Rightarrow & 10 y^2-18 y =0 \\ \Rightarrow & y=0, \frac{9}{5} \Rightarrow x =3, \frac{-12}{5} \end{array}$
Clearly, $M\left(\frac{-12}{5}, \frac{9}{5}\right)$
$\therefore \text { Area of } \triangle O A M=\frac{1}{2} \times\left|\begin{array}{ccc} 0 & 0 & 1 \\ 3 & 0 & 1 \\ \frac{-12}{5} & \frac{9}{5} & 1 \end{array}\right|=\frac{27}{10} \text { (Take only the magnitude) }$

Hint

The given ellipse is $\frac{x^2}{4^2}+\frac{y^2}{2^2}=1$
such that $a^2=16$ and $b^2=4$
$\begin{array}{ll} \therefore & e^2=1-\frac{4}{16}=\frac{3}{4} \\ \Rightarrow & e=\frac{\sqrt{3}}{2} \end{array}$
Let $P(4 \cos \theta, 2 \sin \theta)$ be any point on the ellipse, then equation of normal at $P$ is
$\begin{aligned} \quad 4 x \sin \theta-2 y \cos \theta & =12 \sin \theta \cos \theta \\ \Rightarrow \quad \frac{x}{3 \cos \theta}-\frac{y}{6 \sin \theta} & =1 \end{aligned}$
$\therefore Q$, the point where normal at $P$ meets $X$-axis, has coordinates $(3 \cos \theta, 0)$
$\therefore$ Mid-point of $P Q$ is $M\left(\frac{7 \cos \theta}{2}, \sin \theta\right)$
For locus of point $M$ we consider
$x=\frac{7 \cos \theta}{2} \text { and } y=\sin \theta$
$\Rightarrow \cos \theta=\frac{2 x}{7} \text { and } \sin \theta=y \Rightarrow \frac{4 x^2}{49}+y^2=1 \dots(i)$
Also, the latusrectum of given ellipse is
$\begin{aligned} x & = \pm a e= \pm 4 \times \frac{\sqrt{3}}{2}= \pm 2 \sqrt{3} \\ \text { or } \quad x & = \pm 2 \sqrt{3} \dots(ii) \end{aligned}$
Solving Eqs. (i) and (ii), we get
$\begin{gathered} \frac{4 \times 12}{49}+y^2=1 \\ \Rightarrow \quad y^2=\frac{1}{49} \text { or } y= \pm \frac{1}{7} \\ \therefore \text { The required points are }\left( \pm 2 \sqrt{3}, \pm \frac{1}{7}\right) \text {. } \end{gathered}$

Hint

In $\triangle A B C$, given that $\cos B+\cos C=4 \sin ^2 \frac{A}{2}$
$\begin{aligned} & \Rightarrow \quad 2 \cos \frac{B+C}{2} \cos \frac{B-C}{2}-4 \sin ^2 \frac{A}{2}=0 \\ & \Rightarrow \quad 2 \sin \frac{A}{2}\left[\cos \frac{B-C}{2}-2 \sin \frac{A}{2}\right]=0 \\ & \Rightarrow \quad \sin \frac{A}{2}=0 \text { or }\left(\cos \frac{B-C}{2}-2 \cos \frac{B+C}{2}\right)=0 \qquad\left(\because \frac{A}{2}=90^{\circ}-\left(\frac{B+C}{2}\right)\right) \end{aligned}$
But in a triangle $\sin \frac{A}{2} \neq 0$
$\begin{aligned} & \cos \frac{B-C}{2}-2 \cos \frac{B+C}{2}=0 \\ \Rightarrow \quad & \frac{\cos \left(\frac{B+C}{2}\right)}{\cos \left(\frac{B-C}{2}\right)}=\frac{1}{2} \end{aligned}$
Applying componendo and dividendo, we get
$\begin{aligned} & \frac{\cos \left(\frac{B+C}{2}\right)+\cos \left(\frac{B-C}{2}\right)}{\cos \left(\frac{B+C}{2}\right)-\cos \left(\frac{B-C}{2}\right)}=\frac{1+2}{1-2}=-3 \\ \Rightarrow & -\frac{2 \cos \frac{B}{2} \cos \frac{C}{2}}{2 \sin \frac{B}{2} \sin \frac{C}{2}}=-3 \\ \Rightarrow & \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3} \\ \Rightarrow & \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3} \end{aligned}$
$\begin{array}{rlrl} \Rightarrow & \frac{s-a}{s} =\frac{1}{3} \\ & \text { or } & 2 s =3 a \\ \Rightarrow & a+b+c =3 a \\ \text { or } & b+c =2 a \\ \text { i.e. } & A C+A B & =\text { constant } \text { [ } \because \text { base } B C=a \text { is given to be constant] } \end{array}$
$\Rightarrow A$ moves on an ellipse.

Hint

$\begin{aligned} \left(\frac{x}{\sqrt{3}}\right)^2+(y)^2 & =\left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2 t}{1+t^2}\right)^2 \\ & =\frac{\left(1+t^2\right)^2}{\left(1+t^2\right)^2}=1 \\ \Rightarrow \quad \frac{x^2}{3}+y^2 & =1 \end{aligned}$
Which is an ellipse.

Hint

The given ellipse is $\frac{x^2}{4}+\frac{y^2}{1}=1$
So, $A=(2,0)$ and $B=(0,1)$
If $P Q R S$ is the rectangle in which it is inscribed, then
$P=(2,1) \text {. }$
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be the ellipse circumscribing the rectangle $P Q R S$. Then it passes through $P(2,1)$

$\therefore \quad \frac{4}{a^2}+\frac{1}{b^2}=1 \dots(i)$
Also, given that, it passes through $(4,0)$
$\begin{aligned} & \therefore \quad \frac{16}{a^2}+0=1 \\ & \Rightarrow \quad a^2=16 \\ & \Rightarrow \quad b^2=4 / 3 \quad \text { [substituting } a^2=16 \text { in Eq. (i)] } \\ & \therefore \text { The required ellipse is } \frac{x^2}{16}+\frac{y^2}{4 / 3}=1 \\ & \text { or } x^2+12 y^2=16 \\ & \end{aligned}$

Hint

Tangent to $\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$ at the point $(3 \cos \theta, 2 \sin \theta)$ is
$\frac{x \cos \theta}{3}+\frac{y \sin \theta}{2}=1$

As it passes through $(3,4)$, we get
$\begin{array}{ll} & \cos \theta+2 \sin \theta=1 \\ \Rightarrow \quad & 4 \sin ^2 \theta=1+\cos ^2 \theta-2 \cos \theta \\ \Rightarrow \quad & 5 \cos ^2 \theta-2 \cos \theta-3=0 \\ \Rightarrow \quad & \cos \theta=1,-\frac{3}{5} \\ \Rightarrow \quad & \sin \theta=0, \frac{4}{5} \end{array}$
$\therefore$ Required points are $A(3,0)$ and $B\left(\frac{9}{5}, \frac{8}{5}\right)$.

Hint

Let $H$ be orthocentre of $\triangle P A B$, then as $B H \perp A P, B H$ is a horizontal line through $B$.

$\therefore y$-coordinate of $B=8 / 5$
Let $H$ has coordinate $(\alpha, 8 / 5)$
Then, slope of $P H=\frac{\frac{8}{5}-4}{\alpha-3}=\frac{-12}{5(\alpha-3)}$
and slope of $A B=\frac{\frac{8-0}{5}}{-\frac{9}{5-3}}=\frac{8}{-24}=\frac{-1}{3}$
But $P H \perp A B$
$\begin{array}{ll} \Rightarrow & \frac{-12}{5(\alpha-3)} \times\left(\frac{-1}{3}\right)=-1 \\ \Rightarrow & 4=-5 \alpha+15 \text { or } \alpha=11 / 5 \end{array}$
Hence $H\left(\frac{11}{5}, \frac{8}{5}\right)$.

Hint

Clearly, the moving point traces a parabola with focus at $P(3,4)$ and directrix as
$\begin{aligned} & A B: \frac{y-0}{x-3}=\frac{-1}{3} \\ & x+3 y-3=0 \end{aligned}$
$\therefore$ Equation of parabola is
$(x-3)^2+(y-4)^2=\frac{(x+3 y-3)^2}{10}$
$9 x^2+y^2-6 x y-54 x-62 y+241=0$

Hint

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It passes through $(-3,1)$, so
$\frac{9}{a^2}+\frac{1}{b^2}=1 \dots(i)$
Also,
$b^2=a^2(1-2 / 5)$
$\Rightarrow$ $5 b^2=3 a^2 \dots(ii)$
Solving Eqs. (i) and (ii), we get
$a^2=\frac{32}{3}, b^2=\frac{32}{5}$
So, the equation of the ellipse is $3 x^2+5 y^2=32$

Hint

As rectangle $A B C D$ circumscribed the ellipse
$\frac{x^2}{9}+\frac{y^2}{4}=1$
$\therefore \quad A=(3,2)$
Let the ellipse circumscribing the rectangle $A B C D$ is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Given that, it passes through $(0,4)$
$\therefore \quad b^2=16$
Also, it passes through $A(3,2)$
$\begin{array}{ll} \therefore & \frac{9}{a^2}+\frac{4}{16}=1 \Rightarrow a^2=12 \\ \therefore & e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \end{array}$

Hint

Given, equation of ellipse is $2 x^2+y^2=4$
$\Rightarrow \quad \frac{2 x^2}{4}+\frac{y^2}{4}=1 \Rightarrow \frac{x^2}{2}+\frac{y^2}{4}=1$
Equation of tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{4}=1$ is
$y=m x \pm \sqrt{2 m^2+4} \dots(i)$
$\because$ equation of tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $y=m x+c$, where $c= \pm \sqrt{a^2 m^2+b^2}$
Now, equation of tangent to the parabola
$\because y^2=16 \sqrt{3} x \text { is } y=m x+\frac{4 \sqrt{3}}{m} \dots(ii)$
( $\because$ Equation of tangent to the parabola $y^2=4 a x$ is $\left.y=m x+\frac{a}{m}\right)$
On comparing Eqs. (i) and (ii), we get
$\frac{4 \sqrt{3}}{m}= \pm \sqrt{2 m^2+4}$
Squaring on both the sides, we get
$\begin{aligned} & 16(3)=\left(2 m^2+4\right) m^2 \\ & \Rightarrow \quad 48=m^2\left(2 m^2+4\right) \\ & \Rightarrow \quad 2 m^2+4 m^2-48=0 \\ & \Rightarrow \quad m^4+2 m^2-24=0 \\ & \Rightarrow \quad\left(m^2+6\right)\left(m^2-4\right)=0 \\ & \Rightarrow \quad m^2=4 \quad\left(\because m^2 \neq-6\right) \\ & \Rightarrow \quad m= \pm 2 \\ & \Rightarrow \text { Equation of common tangents are } y= \pm 2 x \pm 2 \sqrt{3} \\ & \end{aligned}$
Thus, statement I is true. statement II is obviously true.

Hint

Equation of circle is $(x-1)^2+y^2=1$
$\Rightarrow$ Radius $=1$ and Diameter $=2$
$\therefore$ Length of semi-minor axis is 2 .
Equation of circle is $x^2+(y-2)^2=4=(2)^2$
$\Rightarrow$ Radius $=2$ and Diameter $=4$
$\Rightarrow$ Length of semi-major axis is 4
We know, equation of ellipse is given by
$\frac{x^2}{\text { (Major-axis) }^2}+\frac{y^2}{(\text { Major-axis) })^2}=1$
$\begin{aligned} \Rightarrow \quad \frac{x^2}{(4)^2}+\frac{y^2}{(2)^2}=1 \Rightarrow \frac{x^2}{16}+\frac{y^2}{4} & =1 \\ \Rightarrow \quad x^2+4 y^2 & =16 \end{aligned}$

Hint

From the given equation of ellipse, we have
$\begin{aligned} & \quad a=4, b=3 e=\sqrt{1-\frac{9}{16}} \\ & \Rightarrow \quad e=\frac{\sqrt{7}}{4} \end{aligned}$

Now, radius of this circle $=a^2=16$
$\Rightarrow \quad \text { Foci }=( \pm \sqrt{7}, 0)$
Now, equation of circle is $(x-0)^2+(y-3)^2=16$
$x^2+y^2-6 y-7=0$

Hint

Vertical line $x=h$, meets the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at
$P\left(h, \frac{\sqrt{3}}{2} \sqrt{\left(4-h^2\right)}\right) \text { and } Q\left(h, \frac{-\sqrt{3}}{2} \sqrt{\left(4-h^2\right)}\right)$
By symmetry, tangents at $P$ and $Q$ will meet each other at $X$-axis

Tangent at $P$ is $\frac{x h}{4}+\frac{y \sqrt{3}}{6} \sqrt{\left(4-h^2\right)}=1$
Which meets $X$-axis at $R\left(\frac{4}{h}, 0\right)$
$\text { Area of } \begin{aligned} \triangle P Q R & =\frac{1}{2} \times \sqrt{3} \sqrt{\left(4-h^2\right)} \times\left(\frac{4}{h}-h\right) \\ \text { i.e. } \quad \Delta(h) & =\frac{\sqrt{3}}{2} \frac{\left(4-h^2\right)^{3 / 2}}{h} \\ \frac{d \Delta}{d h} & =-\sqrt{3}\left[\frac{\sqrt{\left(4-h^2\right)}\left(h^2+2\right)}{h^2}\right]<0 \end{aligned}$
$\therefore \Delta(h)$ is a decreasing function.
$\begin{array}{ll} \therefore & \frac{1}{2} \leq h \leq 1 \\ \Rightarrow & \Delta_{\max }=\Delta\left(\frac{1}{2}\right) \text { and } \Delta_{\max }=\Delta(1) \end{array}$
$\begin{aligned} & \therefore \quad \Delta_1=\frac{\sqrt{3}}{2}\left(\frac{\left(4-\frac{1}{4}\right)^{3 / 2}}{\frac{1}{2}}\right)=\frac{45}{8} \sqrt{5} \\ & \Delta_2=\frac{\sqrt{3}}{2} \cdot \frac{3 \sqrt{3}}{1}=\frac{9}{2} \\ & \therefore \quad \frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=45-36=9 \\ & \end{aligned}$

Hint

Given, equation of ellipse can be written as
$\Rightarrow \quad \begin{aligned} & \frac{x^2}{6}+\frac{y^2}{2}=1 \\ & \Rightarrow \quad a^2=6, b^2=2 \end{aligned}$
Now, equation of any variable tangent is
$y=m x \pm \sqrt{a^2 m^2+b^2} \dots(i)$
where $m$ is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
$y=\frac{-x}{m} \dots(ii)$
Eliminating $m$,we get
$\begin{aligned} & & \left(x^4+y^4+2 x^2 y^2\right) & =a^2 x^2+b^2 y^2 \\ \Rightarrow & & \left(x^2+y^2\right)^2 & =a^2 x^2+b^2 y^2 \\ \Rightarrow & & \left(x^2+y^2\right)^2 & =6 x^2+2 y^2 \end{aligned}$

Hint

The end point of latusrectum of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in first quadrant is $\left(a e, \frac{b^2}{a}\right)$ and the tangent at this point intersects $X$-axis at $\left(\frac{a}{e}, 0\right)$ and $Y$ – axis at $(0, a)$.
The given ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$
Then, $a^2=9, b^2=5$
$\Rightarrow \quad e=\sqrt{\left(1-\frac{5}{9}\right)}=\frac{2}{3}$
$\therefore$ End point of latusrectum in first quadrant is $L(2,5 / 3)$
Equation of tangent at $L$ is $\frac{2 x}{9}+\frac{y}{3}=1$
It meets $X$-axis at $A(9 / 2,0)$ and $Y$-axis at $B(0,3)$
$\therefore \text { Area of } \triangle O A B=\frac{1}{2} \times \frac{9}{2} \times 3=\frac{27}{4}$

By symmetry area of quadrilateral
$=4 \times(\text { Area } \triangle O A B)=4 \times \frac{27}{4}=27 \text { sq units }$

Hint

Let $E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$ and $E_2: \frac{x^2}{c^2}+\frac{y^2}{d^2}=1$, where $c<d$
Also, $S: x^2+(y-1)^2=2$
Tangent at $P\left(x_1, y_1\right)$ to $S$ is $x+y=3$
To find point of contact put $x=3-y$ in $S$, we get $P(1,2)$ writing equation of tangent in parametric form
$\begin{aligned} & \frac{x-1}{\frac{-1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}= \pm \frac{2 \sqrt{2}}{3} \\ & x=\frac{-2}{3}+1 \text { or } \frac{2}{3}+1 \text { and } y=\frac{2}{3}+2 \text { or } \frac{-2}{3}+2 \\ \Rightarrow \quad & x=\frac{1}{3} \text { or } \frac{5}{3} \text { and } y=\frac{8}{3} \text { or } \frac{4}{3} \\ \therefore \quad & Q\left(\frac{5}{3}, \frac{4}{3}\right) \text { and } R\left(\frac{1}{3}, \frac{8}{3}\right) \end{aligned}$
Equation of tangent to $E_1$ at $Q$ is $\frac{5 x}{3 a^2}+\frac{4 y}{3 b^2}=1$ which is identical to $\frac{x}{3}+\frac{y}{3}=1$ $\Rightarrow a^2=5$ and $b^2=4 \Rightarrow e_1^2=1-\frac{4}{5}=\frac{1}{5}$
Equation of tangent to $E_2$ at $R$ is
$\begin{aligned} & \frac{x}{3 c^2}+\frac{8 y}{3 d^2}=1 \text { identical to } \frac{x}{3}+\frac{y}{3}=1 \\ & \Rightarrow \quad c^2=1, d^2=8 \Rightarrow e_2^2=1-\frac{1}{8}=\frac{7}{8} \\ & \therefore \quad e_1^2+e_2^2=\frac{43}{40}, e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}},\left|e_1^2-e_2^2\right|=\frac{27}{40} \end{aligned}$

Hint

$\begin{aligned} & \text { Ellipse } \frac{x^2}{9}+\frac{y^2}{5}=1 \Rightarrow a=3, b=\sqrt{5} \text { and } e=\frac{2}{3} \\ & \therefore \quad f_1=2 \text { and } f_2=-2 \\ & P_1: y^2=8 x \text { and } P_2: y^2=-16 x \\ & T_1: y=m_1 x+\frac{2}{m_1} \\ & 0=-4 m_1+\frac{2}{m_1} \Rightarrow m_1^2=\frac{1}{2} \\ & \end{aligned}$
It passes through $(-4,0)$,
$T_2: y=m_2 x-\frac{4}{m_2}$
It passes through $(2,0)$
$\begin{array}{rcrl} & 0=2 m_2-\frac{4}{m_2} \Rightarrow m_2^2=2 \\ \therefore & \frac{1}{m_1^2}+m_2^2=4 \end{array}$

Hint

For ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1, e=\sqrt{\left(1-\frac{8}{9}\right)}=\frac{1}{3}$
$\therefore \quad F_1(-1,0) \text { and } F_2(1,0)$
Parabola with vertex at $(0,0)$ and focus at $F_2(1,0)$ is $y^2=4 x$. Intersection points of ellipse and parabola are $M\left(\frac{3}{2}, \sqrt{6}\right)$ and $N\left(\frac{3}{2},-\sqrt{6}\right)$
For orthocentre of $\Delta F_1 M N$, clearly one altitude is $X$-axis i.e. $y=0$ and altitude from $M$ to $F_1 N$ is
$y-\sqrt{6}=\frac{5}{2 \sqrt{6}}\left(x-\frac{3}{2}\right)$
Putting $y=0$ in above equation, we get $x=-\frac{9}{10}$
Orthocentre $\left(-\frac{9}{10}, 0\right)$

Hint

Tangents to ellipse at $M$ and $N$ are
$\frac{x}{6}+\frac{y \sqrt{6}}{8}=1 \text { and } \frac{x}{6}-\frac{y \sqrt{6}}{8}=1$
their intersection point is $R(6,0)$
Also, normal to parabola at $M\left(\frac{3}{2}, \sqrt{6}\right)$ is
$y-\sqrt{6}=-\frac{\sqrt{6}}{2}\left(x-\frac{3}{2}\right)$
Its intersection with $x$-axis is $Q\left(\frac{7}{2}, 0\right)$
Now, $\quad \operatorname{ar}(\triangle M Q R)=\frac{1}{2} \times \frac{5}{2} \times \sqrt{6}=\frac{5 \sqrt{6}}{4}$
Also, $\operatorname{area}\left(M F_1 N F_2\right)=2 \times$ Area of $\left(F_1 M F_2\right)$
$\begin{aligned} = & 2 \times \frac{1}{2} \times 2 \times \sqrt{6}=2 \sqrt{6} \\ \frac{\operatorname{arca}(\triangle M Q R)}{\operatorname{area}\left(M F_1 N F_2\right)} & =\frac{5 \sqrt{6}}{4 \times 2 \sqrt{6}}=5: 8 \end{aligned}$

Hint

Here, $e=\frac{1}{2}$ and $x=-\frac{a}{e}=-4$
$\therefore \quad a=2$
and $b^2=a^2\left(1-e^2\right)=4(1-1 / 4)=3$
$\therefore$ Equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
$\Rightarrow$ Equation of normal at $(1,3 / 2)$ is
$\frac{4 x}{1}-\frac{3 y}{3 / 2}=4-3$
or $4 x-2 y=1$

Hint

Tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$y=m x \pm \sqrt{a^2 m^2-b^2}$
Given that $y=\alpha x+\beta$ is the tangent of hyperbola
$\begin{aligned} & \Rightarrow & m=\alpha \text { and } a^2 m^2-b^2 & =\beta^2 \\ & & a^2 \alpha^2-b^2 & =\beta^2 \end{aligned}$
Locus is $a^2 x^2-y^2=b^2$ which is hyperbola.

Hint

For the given ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
$\Rightarrow \quad e=\sqrt{\left(1-\frac{16}{25}\right)}=\frac{3}{5}$
$\Rightarrow$ Eccentricity of hyperbola $=\frac{5}{3}$
Let the hyperbola be $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ then
$B^2=A^2\left(\frac{25}{9}-1\right)=\frac{16}{9} A^2$
$\therefore \frac{x^2}{A^2}-\frac{9 y^2}{16 A^2}=1$, As it passes through focus of ellipse i.e. $(3,0)$
$\therefore$ We get $A^2=9 \Rightarrow B^2=16$
$\therefore$ Equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$, focus of hyperbola is $(5,0)$, vertex of hyperbola is $(3,0)$.

Hint

The length of transverse axis $=2 \sin \theta=2 a$
$\Rightarrow \quad a=\sin \theta$
Also for ellipse $3 x^2+4 y^2=12$
$\begin{gathered} \text { or } \frac{x^2}{4}+\frac{y^2}{3}=1, a^2=4, b^2=3 \\ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2} \end{gathered}$
$\therefore$ Focus of ellipse $=\left(2 \times \frac{1}{2}, 0\right) \Rightarrow(1,0)$
As hyperbola is confocal with ellipse, focus of hyperbola $=(1,0)$
$\begin{array}{rlrl} \Rightarrow & & a e & =1 \Rightarrow \sin \theta \times e=1 \\ \Rightarrow & & e & =\operatorname{cosec} \theta \\ & & b^2 & =a^2\left(e^2-1\right) \\ & & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta \end{array}$
$\therefore$ Equation of hyperbola is
$\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1$
or, $x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

Hint

Two branches of hyperbola have no common tangent but have a common normal joining $S S^{\prime}$.

Hint

Given, equation of hyperbola is
$\frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{\sin ^2 \alpha}=1$
Here, $a^2=\cos ^2 \alpha$ and $b^2=\sin ^2 \alpha$
$\begin{array}{rlrl} \because & b^2 & =a^2\left(e^2-1\right) \\ & \therefore & \sin ^2 \alpha & =\cos ^2 \alpha\left(e^2-1\right) \\ & \text { or } & \sin ^2 \alpha+\cos ^2 \alpha & =\cos ^2 \alpha \cdot e^2 \end{array}$
$\begin{aligned} & \text { or } e^2=1+\tan ^2 \alpha=\sec ^2 \alpha \Rightarrow e=\sec \alpha \\ & \therefore \quad a e=\cos \alpha \cdot \frac{1}{\cos \alpha}=1 \end{aligned}$
Coordinates of foci are $( \pm a e, 0)$ i.e. $( \pm 1,0)$
Hence, abscissae of foci remain constant, when $\alpha$ varies.

Hint

The given hyperbola is
$\begin{aligned} x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6 & =0 \\ \Rightarrow\left(x^2-2 \sqrt{2} x+2\right)-2\left(y^2+2 \sqrt{2} y+2\right) & =6+2-4 \end{aligned}$
$\begin{array}{ll} \Rightarrow & (x-\sqrt{2})^2-2(y+\sqrt{2})^2=4 \\ \Rightarrow & \frac{(x-\sqrt{2})^2}{2^2}-\frac{(y+\sqrt{2})^2}{(\sqrt{2})^2}=1 \end{array}$
$\therefore \quad a=2, b=\sqrt{2} \Rightarrow e=\sqrt{\left(1+\frac{2}{4}\right)}=\sqrt{\frac{3}{2}}$
Clearly, $\triangle A B C$ is a right triangle.
$B\left(a e, \frac{b^2}{a}\right)$

$\begin{aligned} \therefore \quad \text { Area of }(\triangle A B C) & =\frac{1}{2} \times A C \times B C \\ & =\frac{1}{2} \times(a e-a) \times \frac{b^2}{a} \end{aligned}$
$\begin{aligned} & =\frac{1}{2}(e-1) b^2=\frac{1}{2}\left(\sqrt{\frac{3}{2}}-1\right) \cdot 2 \\ & =\left(\sqrt{\frac{3}{2}}-1\right) \end{aligned}$

Hint

The given hyperbola is
$x^2-y^2=\frac{1}{2} \dots(i)$
which is rectangular hyperbola
$\therefore \quad e=\sqrt{2}$
Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Its eccentricity $=\frac{1}{\sqrt{2}}$
$\therefore \quad b^2=a^2\left(1-e^2\right)=a^2\left(1-\frac{1}{2}\right)=\frac{a^2}{2}$
So, the equation of ellipse becomes
$x^2+2 y^2=a^2 \dots(ii)$
Let the hyperbola (i) and ellipse (ii) intersect each other at $P\left(x_1, y_1\right)$.
$\therefore$ (Slope of hyperbola (i) at $\left.\left(x_1, y_1\right)\right) \times$ (Slope of ellipse (ii) at $\left(x_1, y_1\right)=-1$
$\Rightarrow \quad \frac{x_1}{y_1} \times \frac{-x_1}{2 y_1}=-1$
or $x_1^2=2 y_1^2 \dots(iii)$
Also $\left(x_1, y_1\right)$ lies on $x^2-y^2=\frac{1}{2}$
$\therefore \quad x_1^2-y_1^2=\frac{1}{2} \dots(iv)$
From Eqs. (iii) and (iv), we get $y_1^2=\frac{1}{2}$ and $x_1^2=1$ and $\left(x_1, y_1\right)$ lies on ellipse $x^2+2 y^2=a^2$
$\therefore \quad x_1^2+2 y_1^2=a^2$
$\Rightarrow \quad 1+1=a^2 \text { or } a^2=2$
$\therefore$ Equation of ellipse is $x^2+2 y^2=2$ whose foci $( \pm 1,0)$.

Hint

The intersection points of given circle
$x^2+y^2-8 x=0 \dots(i)$
and hyperbola
$4 x^2-9 y^2=36 \dots(ii)$
can be obtained by solving these equations substituting value of $y^2$ from Eq. (i) in Eq. (ii), we get
$\begin{aligned} 4 x^2-9\left(8 x-x^2\right) & =36 \\ 13 x^2-72 x-36 & =0 \\ x & =6,-\frac{6}{13} \\ y^2 & =12,-\frac{660}{169} \text { (not possible) } \end{aligned}$
$\therefore A(6,2 \sqrt{3}) \text { and } B(6,-2 \sqrt{3}) \text { are points of intersection. }$
Equation of tangent to hyperbola having slope $m$ is
$y=m x+\sqrt{\left(9 m^2-4\right)} \dots(iii)$
Equation of tangent to circle is
$y=m(x-4)+\sqrt{\left(16 m^2+16\right)} \dots(iv)$
Eqs. (iii) and (iv) will be identical, then
$m=\frac{2}{\sqrt{5}}$
So, equation of common tangents
$\begin{aligned} y & =\frac{2 x}{\sqrt{5}}+\sqrt{\left(\frac{36}{5}-4\right)} \\ \Rightarrow \quad y & =\frac{2 x}{\sqrt{5}}+\frac{4}{\sqrt{5}} \end{aligned}$
or $2 x-\sqrt{5} y+4=0$

Hint

The intersection points of given circle
$x^2+y^2-8 x=0 \dots(i)$
and hyperbola
$4 x^2-9 y^2=36 \dots(ii)$
can be obtained by solving these equations substituting value of $y^2$ from Eq. (i) in Eq. (ii), we get
$\begin{aligned} 4 x^2-9\left(8 x-x^2\right) & =36 \\ 13 x^2-72 x-36 & =0 \\ x & =6,-\frac{6}{13} \\ y^2 & =12,-\frac{660}{169} \text { (not possible) } \end{aligned}$
$\therefore A(6,2 \sqrt{3}) \text { and } B(6,-2 \sqrt{3}) \text { are points of intersection. }$
Equation of circle with $A B$ as its diameter is
$\begin{aligned} & (x-6)(x-6)+(y-2 \sqrt{3})(y+2 \sqrt{3})=0 \\ & \Rightarrow \quad x^2+y^2-12 x+24=0 \\ & \end{aligned}$

Hint

$\because$ line $2 x+y=1$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\begin{aligned} \therefore & (1)^2 =a^2(-2)^2-b^2 \\ \Rightarrow & 4 a^2-b^2 =1 \end{aligned}$
Intersection point of nearest directrix $x=\frac{a}{e}$ and $X$-axis is $\left(\frac{a}{e}, 0\right)$ As, $2 x+y=1$ passes through $\left(\frac{a}{e}, 0\right)$
$\therefore \quad \frac{2 a}{e}+0=1 \Rightarrow a=\frac{e}{2} \dots(ii)$
and $b^2=a^2\left(e^2-1\right)=\frac{e^2}{4}\left(e^2-1\right) \dots(iii)$
Substituting the values of $a$ and $b$ from Eqs. (ii) and (iii) in Eq. (i), then
$\begin{aligned} & e^2-\frac{e^2}{4}\left(e^2-1\right)=1 \\ & \Rightarrow \quad\left(e^2-4\right)\left(e^2-1\right)=0 \\ & \therefore \quad e^2=4, e^2 \neq 1 \quad(\because e>1) \\ & \text { Hence } e=2 \\ & \end{aligned}$

Hint

Equation of normal at $P(6,3)$ is
$\frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2$
$\because$ Normal intersects the $X$-axis at $(9,0)$, then
$\frac{9 a^2}{6}+0=a^2+b^2 \Rightarrow 3 a^2=2 a^2+2 b^2$
or $a^2=2 b^2$
or $a^2=2 a^2\left(e^2-1\right)$
$\therefore \quad e^2=\frac{3}{2}$
Hence, $e=\sqrt{\frac{3}{2}}$

Hint

Given, ellipse is $x^2+4 y^2=4$
or $\quad \frac{x^2}{4}+\frac{y^2}{1}=1$
$\therefore \quad e=\sqrt{\left(1-\frac{1}{4}\right)}=\frac{\sqrt{3}}{2}$
and foci are $( \pm \sqrt{3}, 0)$
$\therefore$ Eccentricity of hyperbola $=\frac{2}{\sqrt{3}}=e_1$
and $b^2=a^2\left(e_1^2-1\right)=a^2\left(\frac{4}{3}-1\right)=\frac{a^2}{3}$
then equation of hyperbola becomes
$x^2-3 y^2=a^2$
which pass through $( \pm \sqrt{3}, 0)$
$\begin{aligned} 3-0 & =a^2 \\ \Rightarrow a^2 & =3 \end{aligned}$
$\therefore$ Equation of hyperbola is
$x^2-3 y^2=3$
and foci of hyperbola are $\left( \pm \sqrt{3} \times \frac{2}{\sqrt{3}}, 0\right)$ i.e., $( \pm 2,0)$

Hint

Equation of tangent at $\left(x_1, y_1\right)$ is
$\frac{x x_1}{9}-\frac{y y_1}{4}=1 \dots(i)$
Equation of line parallel to
$2 x-y=\lambda \dots(ii)$
$\because$ Line (ii) is tangent of $\frac{x^2}{9}-\frac{y^2}{4}=1$, then
$\begin{array}{rlrl} & & \lambda^2 & =9 \times 2^2-4 \\ \therefore \quad & \lambda & = \pm 4 \sqrt{2} \end{array}$
From Eq. (ii), Equation of tangent is
$2 x-y= \pm 4 \sqrt{2} \dots(iii)$
Comparing Eqs. (i) and (iii), we get $\frac{x_1}{18}=\frac{y_1}{4}= \pm \frac{1}{4 \sqrt{2}}$
or $\quad x_1= \pm \frac{9}{2 \sqrt{2}}$ and $y_1= \pm \frac{1}{\sqrt{2}}$
Hence, points of contact of the tangents on the hyperbola are
$\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right) \text { and }\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

Hint

$H: x^2-y^2=1$
$S:$ Circle with centre $N\left(x_2, 0\right)$
Common tangent to $H$ and $S$ at $P\left(x_1, y_1\right)$ is
$x x_1-y y_1=1 \Rightarrow m_1=\frac{x_1}{y_1}$
Also radius of circle $S$ with centre $N\left(x_2, 0\right)$ through point of contact $\left(x_1, y_1\right)$ is perpendicular to tangent.
$\begin{array}{lll} \therefore & m_1 m_2=-1 \Rightarrow \frac{x_1}{y_1} \times \frac{0-y_1}{x_2-x_1}=-1 \\ \Rightarrow & x_1=x_2-x_1 \text { or } x_2=2 x_1 \end{array}$
$M$ is the point of intersection of tangent at $P$ and $X$-axis
$\therefore \quad M\left(\frac{1}{x_1}, 0\right)$
$\because$ Centroid of $\triangle P M N$ is $(\ell, m)$
$\therefore \quad x_1+\frac{1}{x_1}+x_2=3 \ell \text { and } y_1=3 m$
Using $x_2=2 x_1$,
$\begin{array}{ll} \Rightarrow & \frac{1}{3}\left(3 x_1+\frac{1}{x_1}\right)=l \text { and } \frac{y_1}{3}=m \\ \therefore & \frac{d l}{d x_1}=1-\frac{1}{3 x_1^2}, \frac{d m}{d y_1}=\frac{1}{3} \end{array}$
Also, $\left(x_1, y_1\right)$ lies on $H$,
$\begin{aligned} & \therefore \quad x_1^2-y_1^2=1 \text { or } y_1=\sqrt{\left(x_1^2-1\right)} \\ & \therefore \quad m=\frac{1}{3} \sqrt{\left(x_1^2-1\right)} \\ & \therefore \quad \frac{d m}{d x_1}=\frac{x_1}{3 \sqrt{\left(x_1^2-1\right)}} \\ & \end{aligned}$

Hint

$\because \frac{2 b^2}{a}=8$ and $2 b=\frac{1}{2}(2 a e)$
$\Rightarrow \quad b^2=4 a$ and $b=\frac{1}{2} a e$ or $b^2=\frac{1}{4} a^2 e^2$
$\Rightarrow \quad a^2\left(e^2-1\right)=\frac{1}{4} a^2 e^2 \Rightarrow 4 e^2-4=e^2 \quad$ or $e^2=\frac{4}{3}$
$\therefore \quad e=\frac{2}{\sqrt{3}}$
$\Rightarrow b^2=4 a$ and $b=\frac{1}{2} a e$ or $b^2=\frac{1}{4} a^2 e^2$

Hint

Let Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ When it passes through $P(\sqrt{2}, \sqrt{3})$, then $\frac{2}{a^2}-\frac{3}{b^2}=1$
$\begin{aligned} & \Rightarrow \quad 2 b^2-3 a^2=a^2 b^2 \dots(i) \\ & \text { and } a e=2 \Rightarrow a^2+b^2=4 \dots(ii) \end{aligned}$
From Eqs. (i) and (ii), we get $2\left(4-a^2\right)-3 a^2=a^2\left(4-a^2\right)$
$\begin{aligned} & \Rightarrow a^2-9 a^2+8=0 \Rightarrow\left(a^2-1\right)\left(a^2-8\right)=0 \\ & \therefore a^2=1 \quad\left(\because a^2 \neq 8\right) \\ & \end{aligned}$
From Eq. (ii), $b^2=3$
$\therefore$ Equation of Hyperbola is $\frac{x^2}{1}-\frac{y^2}{3}=1$
Equation of tangent at $P(\sqrt{2}, \sqrt{3})$ is $\frac{x \sqrt{2}}{1}-\frac{y \sqrt{3}}{3}=1$
Which is passes through $(2 \sqrt{2}, 3 \sqrt{2})$

Hint

$( a , b , c ) \because 2 x-y+1=0$ is a tangent of $\frac{x^2}{a^2}-\frac{y^2}{16}=1$, then
$(1)^2=a^2(2)^2-16 \text { or }(2 a)^2=(1)^2+(4)^2$
$\therefore 2 a, 4,1$ are the sides of a right angled triangle.

Hint

I. $x^2+y^2=a^2$
Equations of tangents in terms of slopes are
$y=m x+a \sqrt{\left(m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-m a}{\sqrt{\left(m^2+1\right)}}, \frac{a}{\sqrt{\left(m^2+1\right)}}\right)(Q)$
$\therefore I$ (ii) (Q) (Ans. Q.No. 112) (Here, $a=\sqrt{2}, m= \pm 1$ )
II. $x^2+a^2 y^2=a^2$ or $\frac{x^2}{a^2}+\frac{y^2}{1^2}=1$
Equations of tangents in terms of slopes are
$y=m x+\sqrt{\left(a^2 m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-a^2 m}{\sqrt{\left(a^2 m^2+1\right)}}, \frac{1}{\sqrt{\left(a^2 m^2+1\right)}}\right)$
$\therefore$ II (iv) (R) (Ans. Q.No. 111) (Here, $a=2, m=\frac{-\sqrt{3}}{2}$ )
III. $y^2=4 a x$
Equation of tangent in terms of slopes are
$y=m x+\frac{a}{m} \text { or } m y=m^2 x+a$
and points of contact in terms of slopes are
$\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$
$\therefore$ III (i) (P) (Ans. Q.No. 113) (Here, $a=8, m=1$ )

Hint

I. $x^2+y^2=a^2$
Equations of tangents in terms of slopes are
$y=m x+a \sqrt{\left(m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-m a}{\sqrt{\left(m^2+1\right)}}, \frac{a}{\sqrt{\left(m^2+1\right)}}\right)(Q)$
$\therefore I$ (ii) (Q) (Ans. Q.No. 112) (Here, $a=\sqrt{2}, m= \pm 1$ )
II. $x^2+a^2 y^2=a^2$ or $\frac{x^2}{a^2}+\frac{y^2}{1^2}=1$
Equations of tangents in terms of slopes are
$y=m x+\sqrt{\left(a^2 m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-a^2 m}{\sqrt{\left(a^2 m^2+1\right)}}, \frac{1}{\sqrt{\left(a^2 m^2+1\right)}}\right)$
$\therefore$ II (iv) (R) (Ans. Q.No. 111) (Here, $a=2, m=\frac{-\sqrt{3}}{2}$ )
III. $y^2=4 a x$
Equation of tangent in terms of slopes are
$y=m x+\frac{a}{m} \text { or } m y=m^2 x+a$
and points of contact in terms of slopes are
$\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$
$\therefore$ III (i) (P) (Ans. Q.No. 113) (Here, $a=8, m=1$ )

Hint

I. $x^2+y^2=a^2$
Equations of tangents in terms of slopes are
$y=m x+a \sqrt{\left(m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-m a}{\sqrt{\left(m^2+1\right)}}, \frac{a}{\sqrt{\left(m^2+1\right)}}\right)(Q)$
$\therefore I$ (ii) (Q) (Ans. Q.No. 112) (Here, $a=\sqrt{2}, m= \pm 1$ )
II. $x^2+a^2 y^2=a^2$ or $\frac{x^2}{a^2}+\frac{y^2}{1^2}=1$
Equations of tangents in terms of slopes are
$y=m x+\sqrt{\left(a^2 m^2+1\right)}$
and points of contact in terms of slopes are
$\left(\frac{-a^2 m}{\sqrt{\left(a^2 m^2+1\right)}}, \frac{1}{\sqrt{\left(a^2 m^2+1\right)}}\right)$
$\therefore$ II (iv) (R) (Ans. Q.No. 111) (Here, $a=2, m=\frac{-\sqrt{3}}{2}$ )
III. $y^2=4 a x$
Equation of tangent in terms of slopes are
$y=m x+\frac{a}{m} \text { or } m y=m^2 x+a$
and points of contact in terms of slopes are
$\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$
$\therefore$ III (i) (P) (Ans. Q.No. 113) (Here, $a=8, m=1$ )

Hint

Let $P(x, y)$ be any point on the parabola whose focus is $S(-3,2)$ and the directrix $x+y-4=0$. Draw $P M$ perpendicular to $x+y-4=0$. Then,
$\begin{array}{ll} & S P=P M \\ \Rightarrow \quad & S P^2=P M^2 \\ \Rightarrow \quad & (x+3)^2+(y-2)^2=\left|\frac{x+y-4}{\sqrt{1+1}}\right|^2 \\ \Rightarrow \quad & 2\left(x^2+y^2+6 x-4 y+13\right)=\left(x^2+y^2+16+2 x y-8 x-8 y\right) \\ \Rightarrow \quad & x^2+y^2-2 x y+20 x+10=0 \end{array}$

Hint

Let $P(x, y)$ be any point on the parabola having its focus at $S(-3,0)$ and directrix as the line $x+5=0$. Then,
$S P=P M$, where $P M$ is the length of the perpendicular from $P$ on the directrix
$\Rightarrow \quad S P^2=P M^2$
$\Rightarrow \quad(x+3)^2+(y-0)^2=\left|\frac{x+0 y+5}{\sqrt{1}+0}\right|^2$
$\Rightarrow \quad y^2=4 x+16$, which is the required equation of the parabola.

Hint

The given equation is
$\begin{array}{ll} & 4 y^2+12 x-12 y+39=0 \\ \Rightarrow \quad & 4 y^2-12 y=-12 x-39 \\ \Rightarrow \quad & 4\left(y^2-3 y\right)=-12 x-39 \\ \Rightarrow \quad & 4\left(y^2-3 y+\frac{9}{4}\right)=-12 x-39+9 \end{array}$
$\begin{aligned} & \Rightarrow \quad 4\left(y-\frac{3}{2}\right)^2=-12\left(x+\frac{5}{2}\right) \\ & \Rightarrow \quad\left(y-\frac{3}{2}\right)^2=-3\left(x+\frac{5}{2}\right) \dots(i) \end{aligned}$
Shifting the origin to the point $(-5 / 2,3 / 2)$ without rotating the axes and denoting the new coordinates with respect to these axes by $X$ and $Y$, we obtain
$x=X+\left(-\frac{5}{2}\right), y=Y+\frac{3}{2} \dots(ii)$
Using these relations equation (i), reduces to
$Y^2=-3 X \dots(iii)$
This is of the form $Y^2=-4 a X$. On comparing, we get: $a=\frac{3}{4}$.
Vertex: The coordinates of the vertex with respect to new axes are $(X=0, Y=0)$. So, coordinates of the vertex with respect to old axes are $(-5 / 2,3 / 2) \text { [Putting } X=0, Y=0 \text { in (ii)] }$.
Focus: The coordinates of the focus of the parabola with respect to new axes are
$\left(X=-\frac{3}{4}, Y=0\right) \text {. }$
So, coordinates of the focus with respect to old axes are
$\left(-\frac{13}{4}, \frac{3}{2}\right) \quad\left[\text { Putting } X=-\frac{3}{4}, Y=0 \text { in (ii) }\right]$
Directrix: The equation of the directrix of the parabola with respect to new axes is $X=\frac{3}{4}$. So, equation of the directrix of the parabola with respect to old axes is
$x=-\frac{7}{4} \text { [Putting } X=3 / 4 \text { in (ii)] }$

Hint

(c) Equation of tangent at $(1,7)$ to $x^2=y-6$ is

$2 x-y+5=0$
Now, perpendicular from centre $O(-8,-6)$ to $2 x-y+5=0$ should be equal to radius of the circle
$\begin{aligned} & \therefore\left|\frac{-16+6+5}{\sqrt{5}}\right|=\sqrt{64+36- C } \\ & \quad \Rightarrow \sqrt{5}=\sqrt{100- c } \\ & \quad \Rightarrow c=95 \end{aligned}$

Hint

Given circle is:
$x^2+y^2+2 x-4 y-4=0$
its centre is $(-1,2)$ and radius is 3 units.
Let $A=(x, y)$ be the centre of the circle $C$
$\therefore \frac{x-1}{2}=2 \Rightarrow x=5 \text { and } \frac{y+2}{2}=2 \Rightarrow y=2$
So the centre of $C$ is $(5,2)$ and its radius is 3
$\therefore$ equation of centre $C$ is:
$x^2+y^2-10 x-4 y+20=0$
$\therefore$ The length of the intercept it cuts on the $x$-axis
$=2 \sqrt{g^2-c}=2 \sqrt{25-20}=2 \sqrt{5}$

Hint

(c) Equation of the line passing through the points $(2,3)$ and $(4,5)$ is
$y-3=\left(\frac{5-3}{4-2}\right) (x-2) \Rightarrow x-y+1=0 \dots(1)$
Equation of the perpendicular line passing through the midpoint $(3,4)$ is $x+y-7=0 \ldots .$. (2)
Lines (1) and (2) intersect at the center of the circle. So, the center of the circle is $(3,4)$
Therefore, the radius is
$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(2-3)^2+(3-4)^2}=\sqrt{2} \text { units. }$

Hint

(c) As origin is the only common point to $x$-axis and $y$-axis, so, origin is the common vertex
Let the equation of two of parabolas be $y^2=4 a x$ and $x^2=4 b y$
Now latus rectum of both parabolas $=3$
$\therefore 4 a=4 b=3$
$\Rightarrow a=b=\frac{3}{4}$
$\therefore$ Two parabolas are $y^2=3 x$ and $x^2=3 y$
Suppose $y=m x+c$ is the common tangent.
$\therefore y^2=3 x \Rightarrow(m x+c)^2=3 x \Rightarrow m^2 x^2+(2 m c-3) x+c^2=0$
As, the tangent touches at one point only
So, $b^2-4 a c=0$
$\Rightarrow(2 m c-3)^2-4 m^2 c^2=0$
$\Rightarrow 4 m^2 c^2+9-12 m c-4 m^2 c^2=0$
$\Rightarrow c=\frac{9}{12 m}=\frac{3}{4 m} \dots(i)$
$\therefore x^2=3 y \Rightarrow x^2=3(m x+c) \Rightarrow x^2-3 m x-3 c=0$
Again, $b^2-4 a c=0$
$\Rightarrow 9 m^2-4(1)(-3 c)=0$
$\Rightarrow 9 m^2=-12 c \dots(ii)$
Form $(i)$ and (ii)
$m^2=\frac{-4 c}{3}=\frac{-4}{3}\left(\frac{3}{4 m}\right)$
$\Rightarrow m^3=-1 \Rightarrow m=-1 \Rightarrow c=\frac{-3}{4}$
Hence, $y=m x+c=-x-\frac{3}{4}$
$\Rightarrow 4(x+y)+3=0$

Hint

(a) Here, equation of tangent on $C_1$ at $(2,1)$ is:
$\begin{aligned} & 2 x+y-(x+2)-1=0 \\ & \text { Or } x+y=3 \end{aligned}$
If it cuts off the chord of the circle $C_2$ then the equation of the chord is:
$x+y=3$
$\therefore$ distance of the chord from $(3,-2)$ is :
$d=\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}$
Also, length of the chord is $l=4$
$\begin{aligned} \therefore \text { radius of } C_2=r & =\sqrt{\left(\frac{l}{2}\right)^2+d^2} \\ & =\sqrt{(2)^2+(\sqrt{2})^2}=\sqrt{6} \end{aligned}$

Hint

(a) Equation of tangent at $P (16,16)$ is given as:
$x-2 y+16=0$
Slope of $PC \left( m _1\right)=\frac{4}{3}$
Slope of PB $\left(m_2\right)=-2$
Hence, $\tan \theta=\left|\frac{ m _1- m _2}{1+ m _1 \cdot m _2}\right|=\left|\frac{\frac{4}{3}+2}{1-\frac{4}{3} \cdot 2}\right|$
$\Rightarrow \tan \theta=2$

Hint

(a) Equation of the chord of contact $P Q$ is given by:
$T=0$
or $T \equiv y y_1-4\left(x+x_1\right)$, where $\left(x_1, y_1\right) \equiv(-8,0)$
$\therefore$ Equation becomes: $x=8$
& Chord of contact is $x=8$
$\therefore$ Coordinates of point $P$ and $Q$ are $(8,8)$ and $(8,-8)$
and focus of the parabola is $F(2,0)$
$\therefore$ Area of triangle $P Q F=\frac{1}{2} \times(8-2) \times(8+8)=48$ sq. units

Hint

(a) $A =\{( a , b ) \in R \times R : |a -5|<1,| b -5|<1\}$
Let $a-5=x, b-5=y$
Set A contains all points inside $|x|<1,|y|<1$
$B =\left\{( a , b ) \in R \times R : 4( a -6)^2+9( B -5)^2 \leq 36\right\}$
Set $B$ contains all points inside or on
$\frac{(x-1)^2}{9}+\frac{y^2}{4}=1$

$\therefore \quad( \pm 1, \pm 1)$ lies inside the ellipse.
Hence, $A \subset B$.

Hint

(d) Let for ellipse coordinates of focus and vertex are $(a e, 0)$ and $(a, 0)$ respectively.
$\therefore$ Distance between focus and vertex $=a(1-e)=\frac{3}{2}$ (given)
$\Rightarrow a-\frac{3}{2}=a e$
$\Rightarrow a^2+\frac{9}{4}-3 a=a^2 e^2 \dots(i)$
Length of latus rectum $=\frac{2 b^2}{a}=4$
$\Rightarrow b^2=2 a \dots(ii)$
$e^2=1-\frac{b^2}{a^2}$
$\Rightarrow e^2=1-\frac{2 a}{a^2} \text { (from(ii)) }$
(from(ii))
$\Rightarrow e^2=1-\frac{2}{a} \dots(iii)$
Substituting the value of $e^2$ in eq. (i) we get;
$\Rightarrow a^2+\frac{9}{4}-3 a=a^2\left(1-\frac{2}{a}\right)$
$\Rightarrow a=\frac{9}{4}$
$\therefore$ from eq. (iii) we get;
$e^2=1-\frac{2}{a}=1-\frac{8}{9}=\frac{1}{9}$
$\Rightarrow e=\frac{1}{3}$

Hint

(d) Here equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{36}=1$ Now, $PQ$ is the chord of contact
$\therefore \quad$ Equation of $P Q$ is : $\frac{x(0)}{9}-\frac{y(3)}{36}=1$
$\Rightarrow y=-12$

$\therefore \quad$ Area of $\triangle PQT =\frac{1}{2} \times TR \times PQ$
$\because \quad P \equiv(3 \sqrt{5},-12) \quad \therefore TR =3+12=15$,
$\therefore \quad$ Area of $\triangle PQT =\frac{1}{2} \times 15 \times 6 \sqrt{5}=45 \sqrt{5}$ sq. units

Hint

(a) Here, lines are:
$\begin{aligned} & \sqrt{2} x-y+4 \sqrt{2} k=0 \\ & \Rightarrow \sqrt{2} x+4 \sqrt{2} k=y \dots(i) \end{aligned}$
$\text { and } \sqrt{2} k x+k y-4 \sqrt{2}=0 \dots(ii)$
Put the value of $y$ from (i) in (ii) we get;
$\begin{aligned} & \Rightarrow 2 \sqrt{2} k x+4 \sqrt{2}\left(k^2-1\right)=0 \\ & \Rightarrow x=\frac{2\left(1-k^2\right)}{k}, y=\frac{2 \sqrt{2}\left(1+k^2\right)}{k} \\ & \therefore\left(\frac{y}{4 \sqrt{2}}\right)^2-\left(\frac{x}{4}\right)^2=1 \end{aligned}$
$\therefore$ length of transverse axis
$2 a=2 \times 4 \sqrt{2}=8 \sqrt{2}$
Hence, the locus is a hyperbola with length of its transverse axis equal to $8 \sqrt{2}$

Hint

It is given that the radius of circle is minimum and it touches the curve $y=4-x^2$ and the lines,
$y=|x|$
Let us draw $y=4-x^2$ and $y=|x|$
Let us assume that the radius of the circle is $r$. Then from figure it is clear that the center of the circle will be $(0,4-r)$
Since $y=x$ is tangent to the circle, the line obtained by joining the center with the contact point will be perpendicular to the line $y=x$. The length of this perpendicular is equal to the radius of the circle.
We know that the length of perpendicular (d) drawn from point $(\alpha, \beta)$ on the line
$A x+B y+C=0$ is given by
$d=\frac{|A \alpha+B \beta+C|}{\sqrt{A^2+B^2}}$
Hence, we have
$\begin{gathered} \alpha=0 \\ \beta=4-r \\ d=r \\ \therefore r=\frac{|0-(4-r)+0|}{\sqrt{1^2+(-1)^2}} \\ \Rightarrow r=\frac{|-(4-r)|}{\sqrt{2}} \\ \Rightarrow r=\frac{|(4-r)|}{\sqrt{2}} \end{gathered}$
From the figure, it is clear that radius $r$ is less than 4. Therefore,
$\begin{aligned} & |4-r|=4-r \\ & \therefore r=\frac{4-r}{\sqrt{2}} \\ & \Rightarrow \sqrt{2} r+r=4 \\ & \Rightarrow r=\frac{4}{\sqrt{2}+1} \\ & \end{aligned}$
Let is rationalize the number
$r=\frac{4(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{4(\sqrt{2}-1)}{2-1}=4(\sqrt{2}-1)$

Hint

(c) Let $z=x+iy$
$\operatorname{Im}\left[\left(\frac{i x-y-2}{x+(y-1) i}\right)\left(\frac{x-(y-1) i}{x-(y-1) i}\right)\right]+1=0$
On solving, we get:
$\begin{gathered} 2 x^2+2 y^2-y-1=0 \\ \Rightarrow x^2+y^2-1 / 2 y-1 / 2=0 \\ \Rightarrow x^2+\left(y-\frac{1}{4}\right)^2=\frac{9}{16} \\ \Rightarrow \quad r=\frac{3}{4} \end{gathered}$