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 \(A B C D[latex] with its vertices [latex]A(1,1), B(-1,1), C(-1,-1), D(1,-1)\) \(P\) 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

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

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 \(r[latex] be the radius of required circle.
Clearly, in [latex]\Delta C_1 C C_2, C_1 C=C_2 C=r+1\) 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 \(\left(\frac{-5}{2}, 2\right)[latex] and [latex]r=\frac{5}{2}\)
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
Aslo,
\(
\begin{array}{ll}
\text { Aslo, } \quad & C_1\left(\frac{a}{2}, 0\right) C_2(0,0) \\
& r_1=\frac{|a|}{2} r_2=C \\
& C_1 C_2=r_1-r_2=\frac{|a|}{2} \\
\Rightarrow \quad & c-\frac{|a|}{2}=\frac{|a|}{2} \Rightarrow c=|a|
\end{array}
\)
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 \(L[latex] can be
[latex]
\begin{aligned}
& x-\sqrt{3} y=1 \\
& x-\sqrt{3} y=5
\end{aligned}
\)

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 ne \((1,4)\)
\([\because\) 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
\(
\begin{array}{ll}
& (x-3)^2+(y-4)^2=4^2 \\
\text { or } & (x-3)^2+(y+4)^2=4^2 \\
\text { i.e. } & x^2+y^2-6 x+8 y+9=0 \\
\text { or } & x^2+y^2-6 x+8 y+9=0
\end{array}
\)

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

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

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 \((9,6)[latex], then
[latex]
6=9 m-2 m-m^3
\)
\(
\begin{aligned}
& \Rightarrow \quad m^3-7 m+6=0 \\
& \Rightarrow(m-1)(m-2)(m+3)=0 \\
& \Rightarrow \quad m=1,2,-3 \\
&
\end{aligned}
\)
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

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

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 \(H\),
\(
\begin{aligned}
& \therefore \quad x_1^2-y_1^2=1 \text { or } y_1=\sqrt{\left(x_1^2-1\right)} \\
& \therefore \quad m=\frac{1}{3} \sqrt{\left(x_1^2-1\right)} \\
& \therefore \quad \frac{d m}{d x_1}=\frac{x_1}{3 \sqrt{\left(x_1^2-1\right)}} \\
&
\end{aligned}
\)