Past JEE Main Entrance Papers

Hint

\(
\begin{aligned}
& x^2+y^2=4 \\
& C (0,0) \quad r_1=2 \\
& C^{\prime}(2 \lambda, 0) \quad r_2=\sqrt{4 \lambda^2-9} \\
& \left|r_1-r_2\right|<C C^{\prime}<\left|r_1+r_2\right| \\
& \left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9} \\
& 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9}<4 \lambda^2
\end{aligned}
\)
True \(\lambda \in R \ldots .\). (1)
\(
\begin{aligned}
& 4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{4 \lambda^2-9} \\
& 5<4 \sqrt{4 \lambda^2-9} \text { and } \quad \lambda^2 \geq \frac{9}{4} \\
& \frac{25}{16}<4 \lambda^2-9 \quad \lambda \in\left(-\infty,-\frac{3}{2}\right] \cup\left[\frac{3}{2}, \infty\right) \\
& \frac{169}{64}<\lambda^2 \\
& \lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \dots(2)
\end{aligned}
\)
from (1) and (2)
\(
\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \Rightarrow R -\left[-\frac{13}{8}, \frac{13}{8}\right]
\)
as per question \(a=-\frac{13}{8}\) and \(b=\frac{13}{8}\)
\(\therefore \quad\) required point is \((-1,6)\) with satisfies option (4)

Hint

\(
\begin{array}{r}
x^2+y^2=3 \text { and } x^2=2 y \\
y^2+2 y-3=0 \Rightarrow(y+3)(y-1)=0 \\
y=-3 \text { or } y=1 \\
y =1 x =\sqrt{2} \Rightarrow P(\sqrt{2}, 1)
\end{array}
\)
\(p\) lies on the line
\(
\begin{aligned}
& \sqrt{2} x+y=\alpha \\
& \sqrt{2}(\sqrt{2})+1=\alpha \\
& \alpha=3
\end{aligned}
\)
For circle \(C _1\)
\(Q _1\) lies on \(y\) axis
Let \(Q _1(0, \alpha)\) coordinates
\(
R _1=2 \sqrt{3} \text { (Given }
\)
Line \(L\) act as tangent
Apply \(P = r\) (condition of tangency)
\(
\begin{aligned}
& \Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3} \\
& \Rightarrow|\alpha-3|=6 \\
& \alpha-3=6 \quad \text { or } \quad \alpha-3=-6 \\
& \Rightarrow \alpha=9 \quad \alpha=-3
\end{aligned}
\)
\(
\begin{aligned}
& \triangle P Q_1 Q_2=\frac{1}{2}\left|\begin{array}{ccc}
\sqrt{2} & 1 & 1 \\
0 & 9 & 1 \\
0 & -3 & 1
\end{array}\right| \\
& =\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2} \\
& \left(\Delta P Q_1 Q_2\right)^2=72
\end{aligned}
\)

Hint

Eq. of chord whose mid point is \((h, k)\)
\(
\begin{aligned}
& T=S_1 \\
& x h+y k-(y+k)=h^2+k^2-2 k \\
& x h+y k-y=h^2+k^2-k
\end{aligned}
\)
It passes through \((0,0)\)
\(
\therefore h^2+k^2-k=0
\)
Locus: \(x^2+y^2-y=0\)
Locus intersect \(x+y=1\) at \(P\) and \(Q\)
\(
\begin{aligned}
& \therefore y^2-y+(1-y)^2=0 \\
& 2 y^2-3 y+1=0 \\
& (y-1)(2 y-1)=0 \\
& y=1 \text { or } y=\frac{1}{2} \\
& x=0 \text { or } x=\frac{1}{2} \\
& P (0,1) \text { and } Q \left(\frac{1}{2}, \frac{1}{2}\right) \\
& PQ =\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}
\end{aligned}
\)

Hint

The equation of the circle through \((1,0),(0,1)\) and \((0,0)\) is
\(
\begin{aligned}
& ( x -1)( x )+( y -1)( y )=0 \\
& x^2+y^2-x-y=0 \ldots( i )
\end{aligned}
\)
\(
\begin{aligned}
& \text { Satisfy }(2 k, 3 k) \text { in (i) } \\
& (2 k)^2+(3 k)^2-2 k-3 k=0 \\
& 13 k^2-5 k=0 \\
& k =0, k =\frac{5}{13} \\
& \text { hence } k =\frac{5}{13}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& S:(x-\alpha)^2+(y-\beta)^2=50 \\
& CP = r \\
& \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \\
& \Rightarrow(\alpha+\beta)^2=100
\end{aligned}
\)

Hint

Centre of circle is \((1,-1)\)
Equation of \(A B\) is \(7 x-3 y+10=0 \dots(i)\)
Equation of \(CP\) is \(3 x +7 y +4=0 \ldots\) (ii)
Solving (i) and (ii)
\(
\alpha=\frac{-41}{29}, \beta=\frac{1}{29} \quad \therefore 17 \beta-\alpha=2
\)

Hint

If two circles intersect at two distinct points
\(
\begin{aligned}
& \Rightarrow\left| r _1- r _2\right|< C _1 C _2< r _1+ r _2 \\
& |r-2|<\sqrt{9+16}<r+2 \\
& |r-2|<5 \text { and } r+2>5 \\
& -5< r -2<5 \quad r >3 \ldots \ldots .(1)
\end{aligned}
\)
\(
-3<r<7 \dots(2)
\)
From (1) and (2)
\(
3< r <7
\)

Hint

\(
\begin{aligned}
& C_1: x^2+y^2=25, C_2:(x-\alpha)^2+y^2=16 \\
& 5<\alpha<9
\end{aligned}
\)
\(
\begin{aligned}
& \theta=\sin ^{-1}\left(\frac{\sqrt{63}}{8}\right) \\
& \sin \theta=\frac{\sqrt{63}}{8} \\
& \text { Area of } \triangle OAP =\frac{1}{2} \times \alpha\left(\frac{\beta}{2}\right)=\frac{1}{2} \times 5 \times 4 \sin \theta \\
& \Rightarrow \alpha \beta=40 \times \frac{\sqrt{63}}{8} \\
& \alpha \beta=5 \times \sqrt{63} \\
& (\alpha \beta)^2=25 \times 63=1575
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 2 x+3 y=12 \\
& 3 x-2 y=5 \\
& 13 x=39 \\
& x=3, y=2
\end{aligned}
\)
Center of given circle is \((5,-2)\)
Radius \(\sqrt{25+4-13}=4\)
\(\)
\begin{aligned}
& \therefore C M=\sqrt{4+16}=5 \sqrt{2} \\
& \therefore C P=\sqrt{16+20}=6
\end{aligned}
[/latex

Hint

\(
\begin{aligned}
& (y-2)=m(x-8) \\
& \Rightarrow x \text {-intercept } \\
& \Rightarrow\left(\frac{-2}{m}+8\right) \\
& \Rightarrow y \text {-intercept } \\
& \Rightarrow(-8 m +2) \\
& \Rightarrow OA + OB =\frac{-2}{ m }+8-8 m +2 \\
& f ^{\prime}( m )=\frac{2}{ m ^2}-8=0 \\
& \Rightarrow m ^2=\frac{1}{4} \\
& \Rightarrow m =\frac{-1}{2} \\
& \Rightarrow f \left(\frac{-1}{2}\right)=18 \\
& \Rightarrow \text { Minimum }=18
\end{aligned}
\)

Hint

\(
\begin{aligned}
& h =3 \cos \theta \\
& k =\frac{18}{7} \sin \theta \\
& \therefore \text { locus }=\frac{ x ^2}{9}+\frac{49 y ^2}{324}=1 \\
& e =\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}
\end{aligned}
\)

Hint

Equation of chord with given middle point.
\(
\begin{aligned}
& T = S _1 \\
& \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\
& \frac{8 x+5 y}{200}=\frac{8+2}{200} \\
& y=\frac{10-8 x}{5} \ldots( i ) \\
& \frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1 \quad \text { (put in original equation) } \\
& \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\
& 4 x^2-8 x-15=0 \\
& x=\frac{8 \pm \sqrt{304}}{8} \\
& x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Similarly, } y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5} \\
& y _1=\frac{2-\sqrt{304}}{5} ; y _2=\frac{2+\sqrt{304}}{5}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Distance }=\sqrt{\left( x _1- x _2\right)^2+\left( y _1- y _2\right)^2} \\
& =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5}
\end{aligned}
\)

Hint

Putting \(y^2=3 x^2\) in both the conics
We get \(x^2=b\) and \(\frac{b}{16}+\frac{3}{b}=1\)
\(\Rightarrow b =4,12 \quad( b =4\) is rejected because curves coincide \()\)
\(
\therefore b =12
\)
Hence points of intersection are
\(
( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432
\)

Hint

\(
\begin{aligned}
& 2 b = ae \\
& \frac{ b }{ a }=\frac{ e }{2} \\
& e =\sqrt{1-\frac{ e ^2}{4}} \\
& e =\frac{2}{\sqrt{5}}
\end{aligned}
\)

Hint

\(
\left.\begin{array}{l}
A=(10,0) \\
B=\left(0, \frac{50}{7}\right)
\end{array}\right\} P=(3,5)
\)

\(
\begin{gathered}
ae =3 \\
\frac{ a }{ e }=\frac{25}{3} \\
a =5 \\
b =4
\end{gathered}
\)
Length of \(L R=\frac{2 b^2}{a}=\frac{32}{5}\)

Hint

\(
\begin{aligned}
& \text { Slope of axis }=\frac{1}{2} \\
& y-3=\frac{1}{2}(x-2) \\
& \Rightarrow 2 y-6=x-2 \\
& \Rightarrow 2 y-x-4=0 \\
& 2 x+y-6=0 \\
& 4 x+2 y-12=0 \\
& \alpha+1.6=4 \Rightarrow \alpha=2.4 \\
& \beta+2.8=6 \Rightarrow \beta=3.2
\end{aligned}
\)
Ellipse passes through \((2.4,3.2)\)
\(
\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{a^2}+\frac{\left(\frac{32}{10}\right)^2}{b^2}=1
\)
Also \(1-\frac{ b ^2}{ a ^2}=\frac{1}{2}=\frac{ b ^2}{ a ^2}=\frac{1}{2}\)
\(
\Rightarrow a^2=2 b^2
\)
Put in (1) \(\Rightarrow b^2=\frac{328}{25}\)
\(
\Rightarrow\left(\frac{2 b^2}{a}\right)^2=\frac{4 b^2}{a^2} \times b^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}
\)

Hint

\(
\begin{aligned}
& e_h=\sqrt{1+\sin ^2 \theta} \\
& e_c=\sqrt{1-\sin ^2 \theta} \\
& e_h=\sqrt{7} e_c \\
& 1+\sin ^2 \theta=7\left(1-\sin ^2 \theta\right) \\
& \sin ^2 \theta=\frac{6}{8}=\frac{3}{4} \\
& \sin \theta=\frac{\sqrt{3}}{2} \\
& \theta=\frac{\pi}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& e=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow \frac{1}{2}=1-\frac{b^2}{a^2} \\
& \frac{2 b^2}{a}=14 \\
& e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}} \\
& \left(e_H\right)^2=\frac{3}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& H: \frac{x^2}{16}-\frac{y^2}{9}=1 \\
& e_1=\frac{5}{4} \\
& \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5}
\end{aligned}
\)
Also, ellipse is passing through \(( \pm 5,0)\)
\(
\begin{aligned}
& \therefore a=5 \text { and } b=3 \\
& E: \frac{x^2}{25}+\frac{y^2}{9}=1
\end{aligned}
\)
End point of chord are \(\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)\)
\(
\therefore L _{ PQ }=\frac{10 \sqrt{5}}{3}
\)

Hint

LR subtends \(60^{\circ}\) at centre
\(
\begin{aligned}
& \Rightarrow \tan 30^{\circ}=\frac{b^2 / a}{a e}=\frac{b^2}{a^2 e}=\frac{1}{\sqrt{3}} \\
& \Rightarrow e =\frac{\sqrt{3} b ^2}{9}
\end{aligned}
\)
Also, \(e ^2=1+\frac{ b ^2}{9} \Rightarrow 1+\frac{ b ^2}{9}=\frac{3 b ^4}{81}\)
\(
\begin{aligned}
& \Rightarrow b^4=3 b^2+27 \\
& \Rightarrow b^4-3 b^2-27=0 \\
& \Rightarrow b^2=\frac{3}{2}(1+\sqrt{13})
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 1=3, m=2, n=13 \\
& \Rightarrow l^2+m^2+n^2=182
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x^2}{9}-\frac{y^2}{4}=1 \\
& a^2=9, b^2=4 \\
& b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \\
& e^2=1+\frac{4}{9}=\frac{13}{9} \\
& e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 a e=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Area of } \Delta PS _1 S _2=\frac{1}{2} \times \beta \times s _1 S _2=2 \sqrt{13} \\
& \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \\
& \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \\
& \text { Distance of } P \text { from origin }=\sqrt{\alpha^2+\beta^2} \\
& =\sqrt{18+4}=\sqrt{22}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{x^2}{9}+\frac{y^2}{25}=1 \\
& a=3, b=5 \\
& e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm b e)=(0, \pm 4) \\
& \therefore e_H=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}
\end{aligned}
\)
Let equation hyperbola
\(
\begin{aligned}
& \frac{ x ^2}{ A ^2}-\frac{ y ^2}{ B ^2}=-1 \\
& \therefore B \cdot e _{ H }=4 \quad \therefore B =\frac{8}{3} \\
& \therefore A ^2= B ^2\left( e _{ H }^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore A ^2=\frac{80}{9} \\
& \therefore \frac{ x ^2}{\frac{80}{9}}-\frac{ y ^2}{\frac{64}{9}}=-1 \\
& \text { Directrix : } y= \pm \frac{B}{e_{ H }}= \pm \frac{16}{9} \\
& PS = e \cdot PM =\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\
& =7 \sqrt{\frac{2}{5}}-\frac{8}{3} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} \\
& a=5 \quad b^2=\frac{5 \sqrt{2} a}{2} \\
& b^2=a^2\left(1-e^2\right)=\frac{5 \sqrt{2} a}{2} \\
& \Rightarrow a \left(1- e ^2\right)=\frac{5 \sqrt{2}}{2} \\
& \Rightarrow \frac{5}{ e }\left(1- e ^2\right)=\frac{5 \sqrt{2}}{2} \\
& \Rightarrow \sqrt{2}-\sqrt{2} e ^2= e \\
& \Rightarrow \sqrt{2} e ^2+ e -\sqrt{2}=0 \\
& \Rightarrow \sqrt{2} e ^2+2 e – e -\sqrt{2}=0 \\
& \Rightarrow \sqrt{2} e ( e +\sqrt{2})-1(1+\sqrt{2})=0 \\
& \Rightarrow( e +\sqrt{2})(\sqrt{2} e -1)=0 \\
& \therefore e \neq-\sqrt{2} ; e =\frac{1}{\sqrt{2}} \\
& \frac{ x ^2}{ b ^2}-\frac{ y ^2}{ a ^2 b ^2}=1 \quad a =5 \sqrt{2} \\
& b =5 \\
& a^2 b^2=b^2\left(e_1^2-1\right) \Rightarrow e_1^2=51
\end{aligned}
\)

Hint

Equation of normal to parabola
\(
y=m x-2 m-m^3
\)
this normal passing through center of circle \((2,8)\)
\(
\begin{aligned}
& 8=2 m-2 m-m^3 \\
& m=-2
\end{aligned}
\)
So point \(P\) on parabola \(\Rightarrow\left( am ^2,-2 am \right)=(4,4)\)
And \(C =(2,8)\)
\(
\begin{aligned}
& PC =\sqrt{4+16}=\sqrt{20} \\
& d ^2=20
\end{aligned}
\)

Hint

Parabola is \(x^2=8 y\)
Chord with mid point \(\left( x _1, y _1\right)\) is \(T = S _1\)
\(
\begin{aligned}
& \therefore xx _1-4\left( y + y _1\right)= x _1^2-8 y _1 \\
& \therefore\left( x _1, y _1\right)=\left(1, \frac{5}{4}\right) \\
& \Rightarrow x -4\left( y +\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9 \\
& \therefore x -4 y +4=0 \ldots(i)
\end{aligned}
\)
\((\alpha, \beta)\) lies on (i) & also on \(y^2=4 x\)
\(
\begin{aligned}
& \therefore \alpha-4 \beta+4=0 \ldots(i i) \\
& \& \beta^2=4 \alpha \ldots \text { (iii) }
\end{aligned}
\)
Solving (ii) & (iii)
\(
\begin{aligned}
& \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\
& \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\
& \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\
& (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\
& \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\
& =192
\end{aligned}
\)

Hint

\(
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
0 & 0 & 1 \\
x & y & 1 \\
– x & y & 1
\end{array}\right| \\
& \Rightarrow\left|\frac{1}{2}( xy + xy )\right|=| xy | \\
& \text { Area }(\Delta)=| xy |=\left| x \left(-2 x ^2+54\right)\right| \\
& \frac{ d (\Delta)}{ dx }=\left|\left(-6 x ^2+54\right)\right| \Rightarrow \frac{ d \Delta}{ dx }=0 \text { at } x =3 \\
& \text { Area }=3(-2 \times 9+54)=108
\end{aligned}
\)

Hint

Circle will touch internally.
\(
\begin{aligned}
& C_1 C_2=\left|r_1-r_2\right| \\
& \sqrt{h^2+k^2}=3-\sqrt{h^2+k^2} \\
& \Rightarrow 2 \sqrt{h^2+k^2}=3 \\
& \Rightarrow x^2+y^2=\frac{9}{4}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Area }=2\left[\int_0^1 2 \sqrt{x} d x+\int_1^{\sqrt{5}} \sqrt{5-x^2} d x\right] \\
& =2\left[\left(\frac{4}{3} x^{\frac{3}{2}}\right)_0^1+\left(\frac{x}{2} \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1}\left(\frac{x}{\sqrt{5}}\right)\right)_1^{\sqrt{5}}\right] \\
& =2\left[\frac{4}{3}+\frac{5 \pi}{4}-1-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right] \\
& =\left(\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}}\right) \text { sq. unit }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y d y+3 d x=2 d y \\
& \Rightarrow \quad \frac{y^2}{2}+3 x=2 y+\left.c\right|_{(1,0)} \\
& \Rightarrow \quad c=3 \\
& \therefore \quad(y-2)^2=-6\left(x-\frac{5}{3}\right) \\
& \therefore \text { Vertex }=\left(\frac{5}{3}, 2\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& f(x)=x^2-8 \\
& g(x)=\frac{x}{x-9} \\
& \Rightarrow f(g(10))=f(10)=a=92
\end{aligned}
\)
\(
\begin{aligned}
& g(f(3))=g(1)=b=-\frac{1}{8} \\
& \Rightarrow \text { conic : } \\
& \frac{x^2}{92}+\frac{y^2}{1 / 8}=1 \\
& I( L \cdot R )=\frac{2(1 / 8)}{\sqrt{92}}=\frac{1}{4 \sqrt{92}} \Rightarrow I^2=\frac{1}{16(92)} \\
& e^2=1-\frac{1 / 8}{92}=\frac{92 \times 8-1}{92 \times 8} \\
& \Rightarrow 92 I^2+46 e^2=\frac{1}{16}+\frac{735}{16} \\
& =\frac{736}{16}=46
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Area }=\int_0^{\frac{3}{\sqrt{2}}} x d x+\int_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \sqrt{\frac{18-x^2}{3}} d x \\
& =\frac{1}{2}\left(x^2\right)_0^{\frac{3}{\sqrt{2}}}+\frac{1}{\sqrt{3}}\left[\frac{x}{2} \sqrt{18-x^2}+9 \sin ^{-1}\left(\frac{x}{3 \sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}}
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{2}\left(\frac{9}{2}\right)+\frac{1}{\sqrt{3}}\left[9 \sin ^{-1}(1)-\frac{3}{2 \sqrt{2}} \frac{3 \sqrt{3}}{\sqrt{2}}-9 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\
& =\frac{9}{4}+\frac{1}{\sqrt{3}}\left(\frac{9 \pi}{2}-\frac{9 \sqrt{3}}{4}-\frac{9 \pi}{6}\right) \\
& =\sqrt{3} \pi
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \Rightarrow A B=15 \\
& \left(3 t^2-\frac{3}{t^2}\right)^2+\left(6 t+\frac{6}{t}\right)^2=225 \\
& \Rightarrow 9\left(t^2-\frac{1}{t^2}\right)^2+36\left(t+\frac{1}{t}\right)^2=225 \\
& \Rightarrow 9\left(t+\frac{1}{t}\right)^2\left[\left(t-\frac{1}{t}\right)^2+4\right]=225 \\
& \Rightarrow 9\left(t+\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2=225 \\
& \Rightarrow t+\frac{1}{t}=\left(\frac{225}{9}\right)^{1 / 4}=(25)^{1 / 4}=\sqrt{5} \\
& \text { Equation of } A B \equiv(y-0)=\frac{2}{\left(t-\frac{1}{t}\right)}(x-3) \Rightarrow\left|t-\frac{1}{t}\right|=1 \\
& \Rightarrow y=2 x-6 \Rightarrow y-2 x+6=0 \\
& \text { Distance from origin } \Rightarrow P=\frac{6}{\sqrt{5}} \Rightarrow 10 P^2=\frac{10 \times 36}{5} \\
& =72
\end{aligned}
\)

Hint

Distance of \((5,3)\) to the line \(y=x+c\) is \(\sqrt{2}\)
\(
\begin{aligned}
& \Rightarrow \frac{|3-5-c|}{\sqrt{2}}=\sqrt{2} \\
& |c+2|=2 \\
& \Rightarrow c=0 \\
& \quad c=-4
\end{aligned}
\)
So, the lines are \(y=x\) and \(y=x-4\)
Now, solving these lines with the circle
\(
\begin{aligned}
& y=x \text { and } x^2+y^2-10 x-6 y+30=0 \\
& 2 x^2-16 x+30=0 \\
& x^2-8 x+15=0 \\
& x=3, y=3 \\
& x=5, y=5 \\
& y=x-4 \text { and } x^2+y^2-10 x-6 y+30=0 \\
& 2 x^2-24 x+70=0 \\
& x^2-12 x+35=0 \\
& x=5, y=1 \\
& x=7, y=3 \\
& \sum_{i=1}^4 x_i^2+y_i^2=9+9+25+25+25+1+49+9=152
\end{aligned}
\)

Hint

Take point \(P(x, y)\)

\(
\begin{aligned}
& x=\frac{5+12}{9}, y=\frac{10+4}{9} \\
& P=\left(\frac{17}{9}, \frac{14}{9}\right) \text { (internally) }
\end{aligned}
\)
for externally division.
\(
\begin{aligned}
& x=-\frac{7}{9}, y=\frac{6}{9} \\
& P^{\prime}=\left(\frac{-7}{9}, \frac{6}{9}\right)
\end{aligned}
\)
Locus of \(P\) is the circle whose diameter is \(P P^{\prime}\)
\(
\begin{aligned}
& \left(x-\frac{-17}{9}\right)\left(x+\frac{7}{9}\right)+\left(y-\frac{14}{9}\right)\left(y-\frac{6}{9}\right)=0 \\
& (9 x-17)(9 x+7)+(9 y-14)(9 y-6)=0 \\
& \text { So } 81 x^2-90 x+81 y^2-180 y=35
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|z_1-2 z_2\right|=\left|1-2 \bar{z}_1 z_2\right| \\
& \Rightarrow \quad\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(1-2 \bar{z}_1 z_2\right)\left(1-2 z_1 \bar{z}_2\right)
\end{aligned}
\)
\(
\Rightarrow\left|z_1\right|^2+4\left|z_2\right|^2-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2
\)
\(
=1-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2+4\left|z_1\right|^2\left|z_2\right|^2
\)
\(
\begin{aligned}
\Rightarrow & \left|z_1\right|^2+4\left|z_2\right|^2-4\left|z_1\right|^2\left|z_2\right|^2-1=0 \\
& \left(\left|z_1\right|^2-1\right)\left(4\left|z_2\right|^2-1\right)=0 \\
\Rightarrow & \left|z_1\right|=1 \text { and }\left|z_2\right|=\frac{1}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Slope of BF }=1 \\
& \Rightarrow \text { Slope of } A C \equiv\left(\frac{\alpha-1}{8-5}\right)=-1 \\
& \Rightarrow \alpha-1=-3 \\
& \Rightarrow \alpha=-2
\end{aligned}
\)

From Fig(b)

\((h, k)\) lie on \((y-1)=\frac{-3}{2}(x-6)\)
\(
\begin{aligned}
2 y-2+3 x-18 & =0 \\
2 y+3 x & =20
\end{aligned}
\)
\((h, k)\) lies on circumcircle eg. of circumcircle is \(x^2+y^2=68\)

Hint

Chord with the given middle point is given by \(\Rightarrow\)
\(
\begin{aligned}
T= & S_1 \\
\Rightarrow & y y_1-6\left(x+x_1\right)=y_1^2-12 x_1\left(\left(x_1, y_1\right) \equiv(4,1)\right) \\
& y-6(x+4)=1-48 \\
\Rightarrow & y-6 x+23=0
\end{aligned}
\)
\(\left(\frac{1}{2},-20\right)\) is correct answer.

Hint

Radius of \(C_1=6\)
Radius of \(C_2=2\)
\(
\begin{aligned}
& 2 a e=16 \\
& a e=8 \\
& b^2=a^2 e^2-a^2 \\
& \Rightarrow b^2+a^2=64
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Area }=\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y \\
& =\left[\frac{y^2}{8}+\frac{y}{4}-\frac{y^3}{6}\right]_{-1 / 2}^1 \\
& =\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{6}\right)-\left(\frac{1}{32}-\frac{1}{8}+\frac{1}{48}\right) \\
& =\frac{5}{24}-\left(\frac{3-12+2}{96}\right) \\
& =\frac{5}{24}+\frac{7}{96} \\
& =\frac{27}{96}=\frac{9}{32}
\end{aligned}
\)

Hint

Radius of circle \(=\sqrt{10}\)
\(
\begin{aligned}
& \ln \triangle M P O, \\
& \sqrt{1^2+d^2}=r \\
& 1+d^2=10[\because r=\sqrt{10}] \\
& d^2=9 \\
& d= \pm 3
\end{aligned}
\)
Since, distance is positive, distance between chord is 3 .

Hint

L: \(2 y+x=c\)
\(
y^2=4(x-9)
\)
Now
\(
\begin{aligned}
& \left(\frac{c-x}{2}\right)^2=4(x-9) \\
& x^2-2(c+8) x+c^2+144=0 \\
& D=0 \\
& \Rightarrow c=5 \\
& \therefore L: 2 y+x=5
\end{aligned}
\)
Parabola and \(L\) meets at \((13,-4)\)
Now, distance \(=\sqrt{185}\)

Hint

Given \(f(x)=x^2-5 x \dots(1)\)
\(g(x)=7 x-x^2 \dots(2)\)
Intersection of (1) and (2)
\(
\begin{aligned}
& x^2-5 x=7 x-x^2 \\
& 2 x^2-12 x=0 \\
& 2\left(x^2-6 x\right)=0 \\
& 2 x(x-6)=0 \\
& x=0, x=6 \\
& \therefore A=\int_0^6(g(x)-f(x)) d x \\
& =\int_0^6\left(7 x-x^2-\left(x^2-5 x\right)\right) d x \\
& =\int_0^6\left(7 x-x^2-x^2+5 x\right) d x \\
& =\int_0^6\left(12 x-2 x^2\right) d x \\
& =12\left(\frac{6^2}{2}\right)-\frac{2}{3}(6)^3 \\
& =216-144 \\
& =72 \text { sq. units }
\end{aligned}
\)

Hint

\(\Rightarrow\) Centre \(\equiv(1,2)\)
\(\Rightarrow\) Equation of circle is \((x-1)^2+(y-2)^2=1\)

\(
\begin{aligned}
& P C=5 \\
& C B=1 \\
& \Rightarrow P B \text { = shortest distance }=4
\end{aligned}
\)

Hint

Equation of circle with \(A B\) as diameter
\(
\begin{aligned}
& \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\
& \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0
\end{aligned}
\)
Comparing, \(k=6\)
Latus rectum of ellipse
\(
\begin{aligned}
& x^2+9 y^2=k^2=6^2 \\
& \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \quad \Rightarrow \text { L.R. }=\frac{2 b^2}{a}=\frac{2(4)}{6} \\
& =\frac{4}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Area }=\int_0^2\left(\sqrt{8-x^2}-\sqrt{2 x}\right) d x \\
& =\left[\frac{x}{2} \sqrt{8-x^2}+\frac{8}{2} \sin ^{-1}\left(\frac{x}{2 \sqrt{2}}\right)\right]_0^2-\sqrt{2} \times \frac{2}{3}\left[x^{3 / 2}\right]_0^2 \\
& =2+4 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}) \\
& =\left(\pi-\frac{2}{3}\right) \text { sq. units. }
\end{aligned}
\)

Hint

Base is constant i.e., \(A C=8 \sqrt{3}\)
Area of \(\triangle A B C\) will be maximum when height is maximum
\(\therefore B\) is \((0,0)\)
\(\therefore(\text { Area) })_{\max }==\frac{1}{2} \times 24 \times 8 \sqrt{3}=96 \sqrt{3}\)

Hint

\(
\begin{aligned}
& \frac{h-7}{2}=\frac{k-2}{1}=-2 \frac{(14+2-6)}{5}=-4 \\
& \therefore \quad h=-1, k=-2 \\
& \therefore \quad B^{\prime}(-1,-2) \\
& \therefore \quad A B^{\prime} \equiv \frac{4}{5} y-\frac{3}{5} x+1=0 \\
& a=\frac{4}{5}, b=\frac{-3}{5} \\
& a^2+b^2+3 a b=\frac{16}{25}+\frac{9}{25}-\frac{36}{25}=\frac{-11}{25}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\alpha+2.6}{3}=6, \frac{\beta+\frac{2.15}{2}}{3}=6 \\
& \Rightarrow \alpha=6, \beta=3 \\
& \Rightarrow \text { Also, } \frac{R_1}{R_2}=\frac{2}{1} \\
& R_2=\sqrt{(6-6)^2+\left(6-\frac{15}{2}\right)^2}=\frac{3}{2} \\
& \Rightarrow R_1=2 R_2=3 \\
& \Rightarrow \alpha+\beta+4\left(r_1^2+r_2^2\right)=6+3+4\left(3^2+\frac{9}{4}\right) \\
& =54
\end{aligned}
\)

Hint

In \(\triangle CPB\)
\(
\begin{aligned}
& \cos \frac{\theta}{2}=\frac{P C}{2} \Rightarrow P C=2 \cos \frac{\theta}{2} \\
& \Rightarrow(h-4)^2+(k-5)^2=4 \cos ^2 \frac{\theta}{2} \\
& \text { Now }(x-4)^2+(y-5)^2=\left(2 \cos \frac{\theta}{2}\right)^2 \\
& \Rightarrow r_1=2 \cos \frac{\pi}{6}=\sqrt{3} \\
& r_2=2 \cos \frac{\theta_2}{2} \\
& r_3=2 \cos \frac{\pi}{3}=1 \\
& \Rightarrow r_1^2=r_2^2+r_3^2 \\
& \Rightarrow 3=4 \cos ^2 \frac{\theta_2}{2}+1 \\
& \Rightarrow 4 \cos ^2 \frac{\theta_2}{2}=2 \\
& \Rightarrow \cos ^2 \frac{\theta_2}{2}=\frac{1}{2} \\
& \Rightarrow \theta_2=\frac{\pi}{2}
\end{aligned}
\)

Hint

As \(A\) and \(B\) satisfy both line and circle we have \(\alpha=0 \Rightarrow A(0,0)\) and \(\beta=1\) i.e. \(B(1,1)\)
Centre of circle as \(A B\) diameter is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(=\frac{1}{\sqrt{2}}\)
\(\therefore\) For image of \(\left(\frac{1}{2} ; \frac{1}{2}\right)\) in \(x+y+z\) we get \(\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(3)}{2}\)
\(\Rightarrow\) Image \(\left(-\frac{5}{2},-\frac{5}{2}\right)\)
\(\therefore\) Equation of required circle
\(
\begin{aligned}
& \left(x+\frac{5}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{1}{2} \\
& \Rightarrow x^2+y^2+5 x+5 y+\frac{50}{4}-\frac{1}{2}=0 \\
& \Rightarrow x^2+y^2+5 x+5 y+12=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
& m _{ PQ } \cdot m _{ QR }=-1 \\
& \Rightarrow \frac{10-2}{9+3} \times \frac{10-4}{9-\alpha}=-1 \Rightarrow \alpha=13 \\
& m _{ OP } \cdot m _{ QS }=-1 \Rightarrow m _{ QS }=-\frac{4}{7}
\end{aligned}
\)
Equation of \(QS\)
\(
\begin{aligned}
& y-10=-\frac{4}{7}(x-9) \\
& \Rightarrow 4 x+7 y=106 \ldots .(1) \\
& m_{O R} \cdot m_{R S}=-1 \Rightarrow m_{R S}=-8
\end{aligned}
\)
Equation of \(R S\)
\(
y-4=-8(x-13)
\)
\(
\Rightarrow 8 x+y=108 \ldots .(2)
\)
Solving eq. (1) & (2)
\(
x_1=\frac{25}{2} y_1=8
\)
\(
S\left(x_1, y_1 \right. \text { lies on ) } 2x-ky =1
\)
\(
25-8 k=1
\)
\(
\Rightarrow 8 k =24, k = 3.
\)

Hint

Equation of tangent at \(A(4,-11)\) on circle is
\(
\begin{aligned}
& \Rightarrow 4 x-11 y-3\left(\frac{x+4}{2}\right)+10\left(\frac{y-11}{2}\right)-15=0 \\
& \Rightarrow 5 x-12 y-152=0 \ldots . .(1)
\end{aligned}
\)
Equation of tangent at \(B (8,-5)\) on circle is
\(
\begin{aligned}
& \Rightarrow 8 x-5 y-3\left(\frac{x+8}{2}\right)+10\left(\frac{y-5}{2}\right)-15=0 \\
& \Rightarrow 13 x-104=0 \Rightarrow x=8 \\
& \text { put in }(1) \Rightarrow y=\frac{28}{3} \\
& r=\left|\frac{3.8+\frac{2.28}{3}-34}{\sqrt{13}}\right|=\frac{2 \sqrt{13}}{3}
\end{aligned}
\)

Hint

The given line is polar or \(P (2, \beta)\) w.r.t. given circle
\(
x^2+y^2-4 x-6 y-3=0
\)
Chord or contact
\(
\begin{aligned}
& \alpha x+\beta y-2(x+\alpha)-3(y+\beta)-3=0 \\
& \Rightarrow(\alpha-2) x+(\beta-3) y-(2 \alpha+3 \beta+3)=0 \dots(i)
\end{aligned}
\)
\(\because\) But the equation of chord of contact is given
\(
\text { as : } x+y-3=0 \dots(ii)
\)
comparing the coefficients
\(
\frac{\alpha-2}{1}=\frac{\beta-3}{1}=-\left(\frac{2 \alpha+3 \beta+3}{-3}\right)
\)
On solving \(\alpha=-6\)
\(
\beta=-5
\)
Now \(4 \alpha-7 \beta=11\)

Hint

\(
\begin{aligned}
& (h, k)=\left(\frac{5.5+5(-2)+14 \sqrt{2}}{10+7 \sqrt{2}}, \frac{35+21 \sqrt{2}}{10+7 \sqrt{2}}\right) \\
& h+k=\frac{50+35 \sqrt{2}}{10+7 \sqrt{2}}=5
\end{aligned}
\)

Hint

\(O C \perp C P\) and \(O C \perp C Q\)
\(\Rightarrow P C Q\) is a straight line
\(
O C=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2}=\sqrt{5}
\)
Let \(C P=C Q=I\)
\(
\begin{aligned}
& {[O C P]=\frac{1}{2} \times O C \times I=\frac{\sqrt{35}}{2}} \\
& I=\sqrt{7} \\
& O P=O Q=\sqrt{(O C)^2+I^2}=\sqrt{5+7}=\sqrt{12} \\
& a_1^2+a_2^2+b_1^2+b_2^2=\left(a_1^2+b_2^2\right)+\left(a_2^2+b_2^2\right) \\
& O P^2+O Q^2=12+12=24
\end{aligned}
\)

Hint

\(
\begin{aligned}
& C =(2,3), r =\sqrt{2} \\
& \text { Centre of } G = A =2+4 \frac{1}{\sqrt{2}}, \\
& 3+\frac{4}{\sqrt{2}}=(2+2 \sqrt{2}, 3+2 \sqrt{2}) \\
& A (2+2 \sqrt{2}, 3+2 \sqrt{2}) \\
& B (4+2 \sqrt{2}, 1+2 \sqrt{2}) \\
& \frac{ x -(2+2 \sqrt{2})}{1}=\frac{ y -(3+2 \sqrt{2})}{-1}=2 \\
& \therefore \text { area of trapezium: } \\
& \frac{1}{2}(4+4 \sqrt{2}) 2=4(1+\sqrt{2})
\end{aligned}
\)
\(\therefore\) area of trapezium:

Hint

\(
\begin{aligned}
& x^2+y^2-\frac{(1+a) x}{2}-\frac{(1-a) y}{2}=0 \\
& \text { Centre }\left(\frac{1+a}{4}, \frac{1-a}{4}\right) \Rightarrow(h, k) \\
& P\left(\frac{1+a}{2}, \frac{1-a}{2}\right) \Rightarrow(2 h, 2 k)
\end{aligned}
\)
Equation of chord \(\Rightarrow T = S _1\)
\(
\begin{aligned}
& \Rightarrow( x – y ) \lambda-\frac{2 h ( x +\lambda)}{2}-\frac{(2 k )( y -\lambda)}{2} \\
& =2 \lambda^2-2 h (\lambda)+2 k \lambda
\end{aligned}
\)
Now, \(\lambda(2 h , 2 k )\) satisfies the chord
\(
\begin{aligned}
& \therefore(2 h -2 k ) \lambda- h ( x +\lambda)- k ( y -\lambda) \\
& \Rightarrow 2 \lambda^2+4 k \lambda-4 h \lambda+ h \lambda- k \lambda+ hx + ky =0 \\
& \Rightarrow 2 \lambda^2+\lambda(3 k -3 h )+ ky + hx =0 \\
& \Rightarrow D >0 \\
& \Rightarrow 9( k – h )^2-8( ky + hx )>0 \\
& \Rightarrow 9( k – h )^2-8\left(2 k ^2+2 h ^2\right)>0 \\
& \Rightarrow-7 k ^2-7 h ^2-18 kh >0 \\
& \Rightarrow 7 k ^2+7 h ^2+18 kh <0 \\
& \Rightarrow 7\left(\frac{1- a }{4}\right)^2+7\left(\frac{1+ a }{4}\right)^2+18\left(\frac{1- a ^2}{16}\right)<0
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow 7\left[\frac{2\left(1+a^2\right)}{16}\right]+\frac{18\left(1-a^2\right)}{16}<0, \quad a^2=t \\
& \Rightarrow \frac{7}{8}(1+t)+\frac{18(1-t)}{16}<0 \\
& \Rightarrow \frac{14+14 t+18-18 t}{16}<0 \\
& \Rightarrow 4 t>32 \\
& t>8 \quad a^2>8
\end{aligned}
\)

Hint

Equation of tangent at point \(P(4 \cos \theta, 3 \sin \theta)\) is \(\frac{ x \cos \theta}{4}+\frac{ y \sin \theta}{3}=1\)
So \(A\) is \((4 \sec \theta, 0)\) and point \(B\) is \((0,3 \operatorname{cosec} \theta)\)
\(
\begin{aligned}
& \text { Length } A B =\sqrt{16 \sec ^2 \theta+9 \operatorname{cosec}^2 \theta} \\
& =\sqrt{25+16 \tan ^2 \theta+9 \cot ^2 \theta} \geq 7
\end{aligned}
\)

Hint

Equation of normal of ellipse \(\frac{x^2}{36}+\frac{y^2}{16}=1\) at any point \(P (6 \cos \theta, 4 \sin \theta)\) is
\(3 \sec \theta x-2 \operatorname{cosec} \theta y=10\) this normal is also the normal of the circle passing through the point \((2,0)\) So,
\(6 \sec \theta=10\) or \(\sin \theta=0\) (Not possible)
\(\cos \theta=\frac{3}{5}\) and \(\sin \theta=\frac{4}{5}\) so point \(P =\left(\frac{18}{5}, \frac{16}{5}\right)\)
So the largest radius of circle
\(
r=\frac{\sqrt{320}}{5}
\)
So the equation of circle \((x-2)^2+y^2=\frac{64}{5}\)
Passing it through \((1, \alpha)\)
Then \(\alpha^2=\frac{59}{5}\)
\(
10 \alpha^2=118
\)

Hint

\(
y^2=3 x
\)
Tangent \(P \left( x _1, y _1\right)\) is parallel to \(x +2 y =1\)
Then slope at \(P =-\frac{1}{2}\)
\(
\begin{aligned}
& 2 y \frac{d y}{d x}=3 \\
& \Rightarrow \frac{d y}{d x}=\frac{3}{2 y}=-\frac{1}{2} \\
& \Rightarrow y_1=-3
\end{aligned}
\)
Coordinates of \(P (3,-3)\)
Similarly \(Q\left(\frac{4}{\sqrt{3}}, \frac{1}{\sqrt{5}}\right), R\left(-\frac{4}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)\)
Area of \(\triangle PQR\)
\(
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
3 & -3 & 1 \\
\frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\
-\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1
\end{array}\right| \\
& =\frac{1}{2}\left[3\left(\frac{2}{\sqrt{5}}\right)+3\left(\frac{8}{\sqrt{5}}\right)+0\right]=\frac{30}{2 \sqrt{5}}=3 \sqrt{5}
\end{aligned}
\)

Hint

Equation of normal is
\(
2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^2
\)
Distance from \((0,0)=\frac{4-b^2}{\sqrt{4 \sec ^2 \theta+b^2 \operatorname{cosec}^2 \theta}}\)
Distance is maximum if
\(
\begin{aligned}
& 4 \sec ^2 \theta+b^2 \operatorname{cosec}^2 \theta \text { is minimum } \\
& \Rightarrow \tan ^2 \theta=\frac{b}{2} \\
& \Rightarrow \frac{4-b^2}{\sqrt{4 \cdot \frac{b+2}{2}+b^2 \cdot \frac{b+2}{b}}}=1 \\
& \Rightarrow 4-b^2=b+2 \Rightarrow b=1 \Rightarrow e=\frac{\sqrt{3}}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{a}{e}=8 \ldots \ldots . .(1) \quad a e=2 \dots(ii) \\
& 8 e=\frac{2}{e} \\
& e^2=\frac{1}{4} \Rightarrow e=\frac{1}{2} \\
& a=4 \\
& b^2=a^2\left(1-e^2\right) \\
& =16\left(\frac{3}{4}\right)=12 \\
& \frac{x \cos \theta}{4}+\frac{y \sin \theta}{2 \sqrt{3}}=1 \\
& \sin \theta=\frac{1}{2} \\
& \theta=30^{\circ} \\
& P(2 \sqrt{3}, \sqrt{3}) \\
& Q\left(\frac{8}{\sqrt{3}}, 0\right) \\
& (3 P Q)^2=39
\end{aligned}
\)

Hint

\(
\begin{aligned}
& y^2=24 x \\
& a=6 \\
& x y=2 \\
& A B \equiv t y=x+6 t^2 \dots(1) \\
& A B \equiv T=S_1 \\
& k x+h y=2 h k \dots(2)
\end{aligned}
\)
From (1) and (2)
\(
\begin{aligned}
& \frac{ k }{1}=\frac{ h }{- t }=\frac{2 hk }{-6 t ^2} \\
& \Rightarrow \text { then locus is } y^2=-3 x
\end{aligned}
\)
Therefore directrix is \(4 x=3\)

Hint

Let \(f o g(x)=h(x)\)
\(
\begin{aligned}
& \Rightarrow h^{-1}(x)=\left(\frac{x-7}{2}\right)^{\frac{1}{3}} \\
& \Rightarrow h(x)=f o g(x)=2 x^3+7
\end{aligned}
\)
\(
\begin{aligned}
& f o g(x)=a\left(x^b+c\right)-3 \\
& \Rightarrow a=2, b=3, c=5 \\
& \Rightarrow f o g(a c)=f o g(10)=2007 \\
& g\left(f(x)=(2 x-3)^3+5\right. \\
& \Rightarrow \operatorname{gof}(b)=\operatorname{gof}(3)=32 \\
& \Rightarrow \operatorname{sum}=2039
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 16\left(x^2+4 x\right)-\left(y^2-4 y\right)+44=0 \\
& 16(x+2)^2-64-(y-2)^2+4+44=0 \\
& 16(x+2)^2-(y-2)^2=16 \\
& \frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1
\end{aligned}
\)

\(
\begin{aligned}
& A=\int_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x \\
& A=\int_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}
\end{aligned}
\)
\(
\begin{aligned}
& A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right) \\
& A=\frac{12 \sqrt{6}}{3}+\frac{28}{3} \\
& A=4 \sqrt{6}+\frac{28}{3}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 2 ae =|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2} \\
& ae =\sqrt{2} \\
& a =1 \\
& \Rightarrow b =1 \quad \because e =\sqrt{2} \Rightarrow \text { Hyperbola is rectangular } \\
& \Rightarrow L \cdot R =\frac{2 b ^2}{ a }=2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& 3 x^2-4 y^2=36 , 3 x+2 y=1\\
& m=-\frac{3}{2} \\
& m=+\frac{\sec \theta 3}{\sqrt{12} \cdot \tan \theta} \\
& \Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2} \\
& \sin \theta=-\frac{1}{\sqrt{3}} \\
& (\sqrt{12} \cdot \sec \theta, 3 \tan \theta) \\
& \left(\sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}},-3 \times \frac{1}{\sqrt{2}}\right) \Rightarrow\left(\frac{6}{\sqrt{2}}, \frac{-3}{\sqrt{2}}\right)
\end{aligned}
\)

\(
\sqrt{2}\left( y _0- x _0\right) =-9
\)

Hint

\(
\begin{aligned}
& AB :(\lambda+1) x +\lambda y =4 \\
& AC : \lambda x +(1-\lambda) y +\lambda=0
\end{aligned}
\)
Vertex \(A\) is on \(y\)-axis \(\Rightarrow x=0\)

\(
\begin{aligned}
& \text { So } y=\frac{4}{\lambda}, y=\frac{\lambda}{\lambda-1} \\
& \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} \\
& \Rightarrow \lambda=2
\end{aligned}
\)
\(
\begin{aligned}
& AB : 3 x +2 y =4 \\
& AC : 2 x – y +2=0 \\
& \Rightarrow A (0,2) \text { Let } C (\alpha, 2 \alpha+2)
\end{aligned}
\)
Now (Slope of Altitude through C) \(\left(-\frac{3}{2}\right)=-1\)
\(
\left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2}
\)
So \(C \left(-\frac{1}{2}, 1\right)\)
Let Equation(Fig(b)) of tangent be \(y=m x+\frac{3}{2 m}\)
\(
\begin{aligned}
& m ^2+2 m -3=0 \\
& \Rightarrow m =1,-3
\end{aligned}
\)
So tangent which touches in first quadrant at \(T\) is
\(
\begin{aligned}
& T \equiv\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right) \\
& \equiv\left(\frac{3}{2}, 3\right) \\
& \Rightarrow CT =\sqrt{4+4}=2 \sqrt{2}
\end{aligned}
\)

Hint

\(
\begin{array}{lcl}
\text { Urn A } & \text { Urn B } & \text { Urn C } \\
\end{array}
\)
\(
\begin{array}{llcccc}
\text { Red } & \text { Black } & \text { Red } & \text { Black } & \text { Red } & \text { Black } \\
4 & 6 & 5 & 5 & \lambda & 4
\end{array}
\)
\(
\begin{aligned}
& P\left(\frac{C}{R}\right)=\frac{P(C) P\left(\frac{R}{C}\right)}{P(A) P\left(\frac{R}{A}\right)+P(B) P\left(\frac{R}{B}\right)+P(C) P\left(\frac{R}{C}\right)} \\
& 0.4=\frac{\frac{1}{3} \times \frac{\lambda}{(\lambda+4)}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \frac{\lambda}{(\lambda+4)}} \\
& \Rightarrow \lambda=6
\end{aligned}
\)

\(
\begin{aligned}
& \tan 30^{\circ}=3 t=\frac{3}{2} t^2 \\
& \frac{1}{\sqrt{3}}=\frac{2}{t} \\
& t=2 \sqrt{3}
\end{aligned}
\)
\(
\begin{aligned}
& \left(\frac{3}{2} t ^2, 3 t \right)=(18,6 \sqrt{3}) \\
& \ell^2=18^2+(6 \sqrt{3})^2 \\
& =324+108 \\
& =432
\end{aligned}
\)

Hint

\(
y^2=\frac{x}{2}, T: y=m x+\frac{1}{8 m}
\)
For tangent to \(y^2+1=x\)
\(
\begin{aligned}
& \Rightarrow\left( mx +\frac{1}{8 m }\right)^2+1= x \\
& D =0 \Rightarrow m =\frac{1}{2 \sqrt{2}} \\
& \therefore T : x -2 \sqrt{2} y +1=0 \\
& d =\left|\frac{6+8+1}{\sqrt{9}}\right|=5
\end{aligned}
\)

Hint

Equation of normal
\(
\begin{aligned}
& y=-t x+2 a t+a t^3 \\
& y=-t x+\frac{2}{16} t+\frac{1}{16} t^3
\end{aligned}
\)
It passes through \((0,33)\)
\(
\begin{aligned}
& 33=\frac{t}{8}+\frac{t^3}{16} \\
& t ^3+2 t -528=0 \\
& ( t -8)\left( t ^2+8 t +66\right)=0 \\
& t =8 \\
& P \left( at ^2, 2 at \right)=\left(\frac{1}{16} \times 64,2 \times \frac{1}{16} \times 8\right)=(4,1)
\end{aligned}
\)
Parabola :
\(
\begin{aligned}
& y^2=4(x+y) \\
& \Rightarrow y^2-4 y=4 x \\
& \Rightarrow(y-2)^2=4(x+1)
\end{aligned}
\)
Equation of directix :
\(
\begin{aligned}
& x+1=-1 \\
& x=-2
\end{aligned}
\)
Distance of point \(=6\)

Hint

\(
\begin{aligned}
& y= mx +\frac{4}{ m } \\
& \frac{\left|\frac{4}{ m }\right|}{\sqrt{1+ m ^2}}=2 \sqrt{2} \therefore m = \pm 1 \\
& y = \pm x \pm 4 \text {. Point of contact on parabola }
\end{aligned}
\)
Let \(m =1,\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right)\)
\(R (4,8)\)
Point of contact on circle \(Q (-2,2)\)
\(
\therefore( QR )^2=36+36=72
\)

Hint

\(
\text { As } P ( b , c ) \text { lies on parabola so } c ^2=2 ab \dots(1)
\)
Now equation of tangent to parabola \(y^2=2 a x\) in point form is
\(
y y_1=2 a \frac{\left(x+x_1\right)}{2},\left(x_1, y_1\right)=(b, c)
\)
\(
\Rightarrow y c=a(x+b)
\)
\(
\text { For point } B \text {, put } y=0 \text {, now } x=-b
\)
So, area of \(\triangle PBA , \frac{1}{2} \times AB \times AP =16\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{2} \times 2 b \times c =16 \\
& \Rightarrow bc =16
\end{aligned}
\)
As \(b\) and \(c\) are natural number so possible values of (b, c) are \((1,16),(2,8),(4,4),(8,2)\) and \((16,1)\)
Now from equation (1) \(a=\frac{c^2}{2 b}\) and \(a \in N\), so values of \((b, c)\) are \((1,16),(2,8)\) and \((4,4)\) now values of are 128,16 and 2. Hence sum of values of a is 146.

Hint

\(
\begin{aligned}
& y^2=8 x+4 y+4 \\
& (y-2)^2=8(x+1) \\
& y^2=4 a x \\
& a=2, X=x+1, Y=y-2
\end{aligned}
\)
focus \((1,2)\)
\(
y-2=m(x-1)
\)
Put \((3,0)\) in the above line
\(
m =-1
\)
Length of focal chord \(=16\)

Hint

Let equation of circle is \((x-a)^2+(y-a)^2=a^2\) which is passing through \(P(\alpha, \beta)\)
then \((\alpha-a) 2+(\beta-a)^2=a^2\)
\(
\alpha^2+\beta^2-2 \alpha a-2 \beta \alpha+a^2=0
\)
Here equation of \(AB\) is \(x+y=a\)
Let \(Q \left(\alpha^{\prime}, \beta^{\prime}\right)\) be foot of perpendicular of \(P\) on \(AB\)
\(
\frac{\alpha^{\prime}-\alpha}{1}=\frac{\beta^{\prime}-\beta}{1}=\frac{-(\alpha+\beta-a)}{2}
\)
\(
\begin{aligned}
& PQ ^2=\left(\alpha^{\prime}-\alpha\right)+\left(\beta^{\prime}-\beta\right)=\frac{1}{4}(\alpha+\beta-a)^2+\frac{1}{4}(\alpha+\beta-a)^2 \\
& 121=\frac{1}{2}(\alpha+\beta-a)^2 \\
& 242=\alpha^2+\beta^2-2 \alpha a-2 \beta a+a^2+2 \alpha \beta \\
& 242=2 \alpha \beta \\
& \Rightarrow \alpha \beta=121
\end{aligned}
\)

Hint

Given, The tangents at the points \(P\) and \(Q\) on the circle \(x^2+y^2-2 x+y=5\) meet at the point \(R\left(\frac{9}{4}, 2\right)\), Now plotting the diagram of the above data we have,
Now we know that,
Length of tangent is given by, \(P R=Q R=\sqrt{S_1}\)
So, \(L=\sqrt{S_1}=\sqrt{\left(\frac{9}{4}\right)^2+(2)^2-2 \times \frac{9}{4}+2-5} \Rightarrow L=\frac{5}{4}\)
Now distance between \(C R\) will be, \(C R=\sqrt{\left(\frac{9}{4}-1\right)^2+\left(2+\frac{1}{2}\right)^2}=\sqrt{\frac{25}{16}+\frac{25}{4}}=\sqrt{\frac{125}{16}}=\frac{5 \sqrt{5}}{4}\)
Now finding \(P Q\) by equating area, we get
\(
\begin{aligned}
& \frac{1}{2} \times \frac{5}{2} \times L=\frac{1}{2} \times \frac{P Q}{2} \times C R \\
& \Rightarrow \frac{1}{2} \times \frac{5}{2} \times \frac{5}{4}=\frac{1}{2} \times \frac{P Q}{2} \times \frac{5 \sqrt{5}}{4} \\
& \Rightarrow P Q=\sqrt{5}
\end{aligned}
\)
Now finding area of triangle \(P Q R\) by heron’s formula,
\(
s=\frac{2 \sqrt{5}+5}{4}
\)
So, area will be \(\Delta=\sqrt{s(s-a)(s-b)(s-c)}\)
\(
\begin{aligned}
& \Rightarrow \Delta=\sqrt{\frac{2 \sqrt{5}+5}{4}\left(\frac{2 \sqrt{5}+5}{4}-\frac{5}{4}\right)\left(\frac{2 \sqrt{5}+5}{4}-\frac{5}{4}\right)\left(\frac{2 \sqrt{5}+5}{4}-\sqrt{5}\right)} \\
& \Rightarrow \Delta=\sqrt{\frac{2 \sqrt{5}+5}{4}\left(\frac{2 \sqrt{5}}{4}\right)\left(\frac{2 \sqrt{5}}{4}\right)\left(\frac{5-2 \sqrt{5}}{4}\right)} \\
& \Rightarrow \Delta=\sqrt{\left(\frac{5}{4}\right)\left(\frac{25-20}{16}\right)}=\frac{5}{8}
\end{aligned}
\)

Hint

Given,
\(
C_1: x_2+y_2-4 x-2 y+(5-\alpha)=0
\)
So, its centre will be, \(O_1=(2,1)\) and radius \(=\sqrt{\alpha}\)
And \(C_2: 5 x^2+5 y^2-10 f x-10 g y+36=0\)
\(
\Rightarrow C_2: x^2+y^2-2 f x-2 g y+\frac{36}{5}=0
\)
So, Centre \(O_2=(f, g)\) and radius \(r=\sqrt{f^2+g^2-\frac{36}{5}}\)
Also given \(O_2\) is reflection of \(O_1\) in \(2 x-y+1=0\), so image formula we get,
\(
\begin{aligned}
& \Rightarrow \frac{f-2}{2}=\frac{g-1}{-1}=-2 \cdot\left(\frac{2 \times 2-1+1}{2^2+1^2}\right) \\
& \Rightarrow f=\frac{-6}{5} \text { and } g=\frac{13}{5}
\end{aligned}
\)
So, radius \(r=\sqrt{\left(\frac{-6}{5}\right)^2+\left(\frac{13}{5}\right)^2-\frac{36}{5}}=\frac{\sqrt{25}}{5}=1\)
\(\Rightarrow r=1\) and \(\alpha=1\) as they both are same radius circle,
Hence, \(r+\alpha=2\).

Hint

The given information can be represented in the form of the diagram below.
Since angle in a semicircle is a right angle.
Hence, the other circle will be passing through the centre of the first circle.
The two ends of diameter of the circle is \(O(0,0)\) and \(C(3,-2)\).
Hence, the required equation of circle is \((x-0)(x-3)+(y-0)(y-(-2))=0\)
\(
\Rightarrow x^2+y^2-3 x+2 y=0
\)
Put \(\left(\alpha, \frac{1}{2}\right)\) in the above equation \(\Rightarrow \alpha^2+\frac{1}{4}-3 \alpha+1=0\)
\(
\begin{aligned}
& \Rightarrow \alpha^2-3 \alpha+\frac{5}{4}=0 \\
& \Rightarrow 4 \alpha^2-12 \alpha+5=0 \\
& \Rightarrow 4 \alpha^2-10 \alpha-2 \alpha+5=0 \\
& \Rightarrow 2 \alpha(2 \alpha-5)-1(2 \alpha-5)=0 \Rightarrow \alpha=\frac{1}{2}, \frac{5}{2}
\end{aligned}
\)
Therefore this is the correct option.

Hint

Given equation of ellipse is \(\frac{x^2}{36}+\frac{y^2}{9}=1\)
It is of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
With \(a=6, b=3\).
The centre of the circle is \(C(2,0)\).
Let \(P(a \cos \theta, b \sin \theta)\) be the common point for ellipse and circle.
\(
P \equiv(6 \cos \theta, 3 \sin \theta)
\)
Equation of normal to the ellipse will be \(\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2\).
\(
\Rightarrow 6 x \sec \theta-3 y \operatorname{cosec} \theta=36-9
\)
But the normal through \(P\) passes through the centre of the circle \(C(2,0)\).
\(
\begin{aligned}
& \Rightarrow 6(2) \sec \theta-3(0) \operatorname{cosec} \theta=27 \\
& \Rightarrow \sec \theta=\frac{27}{12} \\
& \Rightarrow \cos \theta=\frac{4}{9} \text { and } \sin \theta=\frac{\sqrt{65}}{9} \\
& P \equiv\left(6 \times \frac{4}{9}, 3 \times \frac{\sqrt{65}}{9}\right) \\
& P \equiv\left(\frac{8}{3}, \frac{\sqrt{65}}{3}\right)
\end{aligned}
\)
Now \(C P=r=\sqrt{\left(2-\frac{8}{3}\right)^2+\left(0-\frac{\sqrt{65}}{3}\right)^2}\)
\(
\begin{aligned}
& \Rightarrow r^2=\frac{69}{9} \\
& \Rightarrow 12 r^2=12 \times \frac{69}{9}=92
\end{aligned}
\)
Therefore, the required answer is 92.

Hint

Assuming a circle of radius \(r\) which touches both the axis is given by, \((x-r)^2+(y-r)^2=r^2\) Now given line \(x+y=2\) makes the intercept 2 the circle, so plotting the diagram we get,

Where \(d=OC\) perpendicular distance of centre from line \(x+y=2\)

Now from diagram,
\(
A C=B C=1, O C=\sqrt{r^2-1}
\)
Now using the formula of perpendicular distance of point from the line we get,
\(
\begin{aligned}
& O C=\left|\frac{2 r-2}{\sqrt{2}}\right|=\sqrt{r^2-1} \\
& \Rightarrow 2(r-1)^2=r^2-1 \\
& \Rightarrow r^2-4 r+3=0 \\
& \Rightarrow r_1=1, r_2=3
\end{aligned}
\)
Hence, \(r_1^2+r_2^2-r_1 r_2=1+9-3=7\)

Hint

Given that \(\frac{x^2}{36}+\frac{y^2}{4}=1\)
Observe that \((3 \sqrt{3}, 1)\) lies on the given ellipse.
Equation of tangent to the ellipse at the given point is \(T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1\)
\(
\Rightarrow \frac{\sqrt{3} x}{12}+\frac{y}{4}=1
\)
Equation of normal passing through same point will be \(N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}}\)
\(
\begin{aligned}
& \Rightarrow \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\
& \Rightarrow 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\
& \Rightarrow 3 x-\sqrt{3} y=8 \sqrt{3}
\end{aligned}
\)
Now tangent and normal intersects the \(y\)-axis at \(A\) and \(B\) respectively.
\(
A(0,4) \quad B(0,-8)
\)
Equation of circle with \(A B\) as diameter will be
\(
\begin{aligned}
& \left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0 \\
& \Rightarrow x^2+(y-4)(y+8)=0
\end{aligned}
\)
Line \(x=2 \sqrt{5}\) intersects the circle.
\(
\begin{aligned}
& \Rightarrow 20+y^2+4 y-32=0 \\
& \Rightarrow y^2+4 y-12=0 \\
& \Rightarrow(y+6)(y-2)=0
\end{aligned}
\)
Hence \(P(2 \sqrt{5},-6) Q(2 \sqrt{5}, 2)\)
Now let us find the equation of tangents to the circle at \(P, Q[latex].
[latex]
\begin{aligned}
& \Rightarrow T: x x_1+y y_1+2 y+2 y_1-32=0 \\
& \Rightarrow T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\
& \Rightarrow 2 \sqrt{5} x-4 y=44
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow T_1: \sqrt{5} x-2 y=22 \ldots \ldots \ldots \text { (i) } \\
& \Rightarrow T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\
& \Rightarrow 2 \sqrt{5} x+4 y=28 \\
& \Rightarrow T_2: \sqrt{5} x+2 y=14 \ldots \ldots \ldots \text { (ii) }
\end{aligned}
\)
From (i) and (ii)
\(
\begin{aligned}
& \alpha=\frac{18}{\sqrt{5}} \beta=-2 \\
& \Rightarrow \alpha^2-\beta^2=\frac{304}{5}
\end{aligned}
\)
Hence this is the required option.

Hint

Given,
The centre of a circle \(C\) be \(\alpha, \beta\) and its radius \(r<8\), And \(3 x+4 y=24\) and \(3 x-4 y=32\) be two tangents
So, the distance from centre will be radius,
\(
\left|\frac{3 \alpha+4 \beta-24}{5}\right|=\left|\frac{3 \alpha-4 \beta-32}{5}\right|
\)
Taking + sign we get,
\(
\begin{aligned}
& \Rightarrow 3 \alpha+4 \beta-24=3 \alpha-4 \beta-32 \\
& \Rightarrow 8 \beta=-8 \Rightarrow \beta=-1
\end{aligned}
\)
And taking – sign we get,
\(
\begin{aligned}
& 3 \alpha+4 \beta-24=-3 \alpha+4 \beta+32 \\
& \Rightarrow 6 \alpha=56 \Rightarrow \alpha=\frac{56}{6}
\end{aligned}
\)
And \(4 x+3 y=1\) be a normal to \(C\)
So, \((\alpha, \beta)\) lies on \(4 x+3 y=1\)
\(
\Rightarrow 4 \alpha+3 \beta=1
\)
So, \(\alpha=1\) when \(\beta=-1\) and \(\beta=\frac{1}{3}\left(1-4 \times \frac{28}{3}\right)=\frac{-109}{9}\), if \(\alpha=\frac{28}{3}\)
Hence, radius will be,
\(
r=\left|\frac{3-4-24}{5}\right|=5<8 \text { if }(\alpha, \beta)=(1,-1)
\)
And if \((\alpha, \beta)=\left(\frac{28}{3}, \frac{-109}{9}\right)\) then \(r>8\), so neglected.
\(
\begin{aligned}
& \therefore r=5, \alpha=1, \beta=-1 \\
& \therefore \alpha-\beta+r=7
\end{aligned}
\)

Hint

If \(S \equiv x^2+y^2+2 g x+2 f y+c=0\).
\(C \equiv(-g,-f)\) and \(r=\sqrt{g^2+f^2-c}\).
The given circles are \(x^2+y^2-18 x-15 y+131=0\)
\(
\Rightarrow C_1\left(9, \frac{15}{2}\right), r_1=\sqrt{81+\frac{225}{4}-131}=\frac{5}{2}
\)
and \(x^2+y^2-6 x-6 y-7=0\)
\(\Rightarrow C_2(3,3), r_2=\sqrt{9+9+7}=5\)
\(\Rightarrow d=C_1 C_2=\sqrt{(9-3)^2+\left(\frac{15}{2}-3\right)^2}=\sqrt{36+\frac{81}{4}}=\frac{15}{2}\)
\(\Rightarrow r_1+r_2=\frac{5}{2}+5=\frac{15}{2}\)
\(\Rightarrow C_1 C_2=r_1+r_2\)
\(\therefore\) Circles touch each other externally, 3 common tangents.
Hence this is the correct option.

Hint

\(
\begin{aligned}
& E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \rightarrow e \\
& H: x^2-y^2=\frac{1}{2} \Rightarrow e^{\prime}=\sqrt{2} \\
& e=\frac{1}{\sqrt{2}} \\
& \because e^2=\frac{1}{2} \\
& 1-\frac{b^2}{a^2}=\frac{1}{2} \Rightarrow \frac{b^2}{a^2}=\frac{1}{2} \\
& a^2=2 b^2
\end{aligned}
\)
\(E \& H\) are at right angle
they are confocal
Focus of Hyperbola = focus of ellipse
\(
\begin{aligned}
& \left( \pm \frac{1}{\sqrt{2}} \cdot \sqrt{2}, 0\right)=\left( \pm \frac{a}{\sqrt{2}}, 0\right) \\
& a=\sqrt{2} \\
& \because a^2=2 b^2 \Rightarrow b^2=1 \\
& \text { Length of } L R=\frac{2 b^2}{a}=\frac{2(1)}{\sqrt{2}} \\
& =\sqrt{2}
\end{aligned}
\)
Square of \(L R=2\)

Hint

Equation of line \(AB\) or \(AP\) is
\(
\begin{aligned}
& \frac{x}{3}+\frac{y}{1}=1 \\
& x+3 y=3 \\
& x=(3-3 y)
\end{aligned}
\)
Intersection point of line AP & circle is \(P \left( x _0, y _0\right)\)
\(
\begin{aligned}
& x^2+y^2=9 \Rightarrow(3-3 y)^2+y^2=9 \\
& \Rightarrow 3^2\left(1+y^2-2 y\right)+y^2=9 \\
& \Rightarrow 5 y^2-9 y=0 \Rightarrow y(5 y-9)=0 \\
& \Rightarrow y=9 / 5
\end{aligned}
\)
\(
\begin{aligned}
& \text { Hence } x=3(1-y)=3\left(1-\frac{9}{5}\right)=3\left(\frac{-4}{5}\right) \\
& x=\frac{-12}{5} \\
& P\left(x_0, y_0\right)=\left(\frac{-12}{5}, \frac{9}{5}\right)
\end{aligned}
\)
Area of \(\triangle AOP\) is \(=\frac{1}{2} \times OA \times\) height
Height \(=9 / 5, O A=3\)
\(
=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{ m }{ n }
\)
Compare both side \(m =27, n =10 \Rightarrow m – n =17\)

Hint

Given,
\(
\begin{aligned}
& 15 x^2+19 y^2=285 \\
& \Rightarrow \frac{x^2}{19}+\frac{y^2}{15}=1
\end{aligned}
\)
Now plotting the diagram with given circle of radius 4 we get,
Now, let the equation of tangent of ellipse be \(y=m x \pm \sqrt{19 m^2+15}\left\{\right.\) as \(\left.c^2=a^2 m^2+b^2\right\}\) If it is tangent to circle \(x^2+y^2=16\) then using the formula, \(c^2=r^2\left(1+m^2\right)\) we get,
\(
\begin{aligned}
& \frac{\sqrt{19 m^2+15}}{\sqrt{1+m^2}}=4 \\
& \Rightarrow 19 m^2+15=16+16 m^2 \\
& \Rightarrow 3 m^2=1 \\
& \Rightarrow m= \pm \frac{1}{\sqrt{3}} \\
& \Rightarrow \theta=30^{\circ} \text { with } x \text {-axis, }
\end{aligned}
\)
Hence, Angle made by tangent with minor axis i.e., with \(y\) axis is \(\frac{\pi}{3}\)

Hint

Given,
\(
\begin{aligned}
& E_k: k x^2+k^2 y^2=1 \\
& \Rightarrow E_k: \frac{x^2}{\left(\frac{1}{\sqrt{k}}\right)^2}+\frac{y^2}{\left(\frac{1}{k}\right)^2}=1
\end{aligned}
\)
Now equation of the chord joining the points \(\left(\frac{1}{\sqrt{k}}, 0\right) \&\left(0, \frac{1}{k}\right)\) will be,
\(
\begin{aligned}
& L_k: \frac{x}{\left(\frac{1}{\sqrt{k}}\right)}+\frac{y}{\left(\frac{1}{k}\right)}=1 \\
& \Rightarrow \sqrt{k} x+k y-1=0
\end{aligned}
\)
Now \(r_k=\) Perpendicular distance of \(L_k\) from \((0,0)\) we get,
\(
\begin{aligned}
& r_k=\left|\frac{-1}{\sqrt{k+k^2}}\right| \\
& \Rightarrow r_k^2=\frac{1}{k+k^2}
\end{aligned}
\)
Now putting the value of \(r_k^2\) in \(\sum_{k=1}^{20} \frac{1}{r_k^2}\) we get,
\(
\begin{aligned}
& \sum_{k=1}^{20} \frac{1}{r_k^2}=\sum_{k=1}^{20} k+k^2=\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\
& =210+2870=3080
\end{aligned}
\)

Hint

Given, \(P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R\) and \(S\) be four points on the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\)
Now, \(O P=r_1=\sqrt{\left(\frac{2 \sqrt{3}}{\sqrt{7}}\right)^2+\left(\frac{6}{\sqrt{7}}\right)^2}=\sqrt{\frac{48}{7}}\) where \(O\) is origin,
Let \(P\) be \(\left(r_1 \cos \theta, r_1 \sin \theta\right)\)
\(P\) lies on ellipse, so we get,
\(
\begin{aligned}
& \frac{r_1{ }^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1 \\
& \Rightarrow \frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48}
\end{aligned}
\)
Let \(R\) be \(\left(-r_2 \sin \theta, r_2 \cos \theta\right)\) as \(P Q \& R S\) are perpendicular and pass through origin,
So, \(\frac{r_2{ }^2 \sin ^2 \theta}{4}+\frac{r_2{ }^2 \cos ^2 \theta}{9}=1\)
\(
\Rightarrow \frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \ldots
\)
Now adding equation (i) & (ii) we get,
\(
\frac{1}{r_2{ }^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}
\)
Now solving,
\(
\begin{aligned}
& \frac{1}{P Q^2}+\frac{1}{R S^2}=\frac{1}{4}\left(\frac{1}{O P^2}+\frac{1}{O R^2}\right) \\
& \Rightarrow \frac{1}{P Q^2}+\frac{1}{R S^2}=\frac{1}{4}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}\right) \\
& \Rightarrow \frac{1}{P Q^2}+\frac{1}{R S^2}=\frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144}=\frac{p}{m} \\
& \therefore p+m=157
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { L.R. }=\frac{2 b^2}{a}=\frac{1}{2} \\
& 4 b^2=a \dots(i) \\
& \text { Ellipse } \frac{(x-1)^2}{a^2}+\frac{y^2}{b^2}=1 \\
& m_{B_1 F_1}=\frac{1}{\sqrt{3}} \\
& \frac{b}{a}=\frac{1}{\sqrt{3}} \\
& 3 b^2=a^2 e^2=a^2-b^2 \\
& 4 b^2=a^2 \dots(ii) \\
& \text { From }(i) \text { and (ii) } \\
& a=a^2 \\
& \therefore a=1 \\
& b^2=\frac{1}{4} \\
& ((2 a)+(2 b))^2=9
\end{aligned}
\)

Hint

Given, Equation of hyperbola,
\(
H_n: \frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n \in N
\)
Now using the eccentricity formula we get,
\(
\begin{aligned}
& e^2=1+\frac{3+n}{1+n} \\
& \Rightarrow e^2=\frac{2 n+4}{n+1}=\frac{2(n+2)}{n+1}
\end{aligned}
\)
Now check when \((n+1)=9,25,49, \ldots \ldots\) is perfect square and then at the same time \(2(n+2)\) should also be perfect square,
So checking for \(n=8 \rightarrow e^2=\frac{20}{9}\)
\(
\begin{aligned}
& n=24 \rightarrow e^2=\frac{52}{25} \\
& n=48 \rightarrow e^2=\frac{100}{49} \Rightarrow e=\frac{10}{7}
\end{aligned}
\)
Hence, for \(n=48\) both \(n+1 \& 2(n+2)\) are perfect square,
Now we know that,
Length of latusrectum is given by \(l=\frac{2 b^2}{a}=\frac{2(n+3)}{\sqrt{n+1}}=\frac{2 \times 51}{7}\)
Hence, \(21 l=\frac{42 \times 51}{7}=306\)

Hint

Given, The tangent to the parabola \(y^2=12 x\) at the point \((3, \alpha)\) be perpendicular to the line \(2 x+2 y=3\)
So, Slope of tangent \(=1=\frac{6}{\alpha} \Rightarrow \alpha=6\)
So, equation of hyperbola will be,
\(
\begin{aligned}
& 36 x^2-9 y^2=324 \\
& \Rightarrow \frac{x^2}{9}-\frac{y^2}{36}=1
\end{aligned}
\)
Now equation of tangent at \((5,8)\) will be,
\(
\begin{aligned}
& \frac{5 x}{9}-\frac{8 y}{36}=1 \\
& \Rightarrow 5 x-2 y=9
\end{aligned}
\)
So, slope of normal \(=\frac{-2}{5}\)
Now finding, equation of normal \(y-8=\frac{-2}{5}(x-5)\)
\(
\begin{aligned}
& \Rightarrow 5 y-40=-2 x+10 \\
& \Rightarrow 5 y+2 x=50
\end{aligned}
\)
Now finding, distance from \((6,-4)\), we get,
\(
\begin{aligned}
& =\left|\frac{12-20-50}{\sqrt{29}}\right|=\frac{58}{\sqrt{29}} \\
& =2 \sqrt{29}
\end{aligned}
\)
Hence, \((2 \sqrt{29})^2=116\)

Hint

Given that \(H: \frac{y^2}{25}-\frac{x^2}{16}=1\)
Equation of tangent to the hyperbola is: \(y=m x+\sqrt{25-16 m^2}\)
Passes through \((4,1)\)
\(
\begin{aligned}
& \Rightarrow y=m x+\sqrt{25-16 m^2} \\
& \Rightarrow 1=4 m+\sqrt{25-16 m^2} \\
& \Rightarrow 4 m^2-m-3 \\
& \Rightarrow m_1=1, m_2=\frac{-3}{4} \\
& \Rightarrow\left|m_1\right|=1,\left|m_2\right|=\frac{3}{4}
\end{aligned}
\)
Equation of tangents with slopes \(1 \& \frac{3}{4}\)
\(
\begin{aligned}
& \Rightarrow y=x-3 \ldots \ldots .( i ) \\
& \Rightarrow y=\frac{3}{4} x-4 \ldots \ldots \text { (ii) } \\
& \text { (i) \& (ii) } \Rightarrow Q \equiv(-4,-7) \\
& P \equiv(4,1) \\
& \Rightarrow(P Q)^2=8^2+8^2=128
\end{aligned}
\)
On converting \((i)\) and \((i i)\) in intercept form we get,
\(
\begin{aligned}
& \text { (i) } \Rightarrow \alpha=3,( ii ) \Rightarrow \beta=\frac{16}{3} \\
& \Rightarrow \frac{(P Q)^2}{\alpha \beta}=\frac{128}{3 \times \frac{16}{3}}=8
\end{aligned}
\)
Hence this is the required option.

Hint

The given equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, a e=2, e=\frac{3}{2}\)
\(
\begin{aligned}
& \Rightarrow a=\frac{4}{3} \\
& \Rightarrow b^2=a^2 e^2-a^2=4-\frac{16}{9}=\frac{20}{9}
\end{aligned}
\)
The tangent is perpendicular to \(2 x+3 y=6\).
Slope of tangent is \(m=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}\)
Equation of tangent, \(m=\frac{3}{2}\)
Equation of tangent is \(y=m x-\sqrt{a^2 m^2-b^2}\)
\(
\Rightarrow y=\frac{3}{2} x-\sqrt{\frac{16}{9} \times \frac{9}{4}-\frac{20}{9}}
\)
\(\left(\because\right.\) Tangent is in \(1^{\text {st }}\) quadrant \(\left.\Rightarrow C<0\right)\)
\(
\Rightarrow y=\frac{3}{2} x-\frac{4}{3}
\)
Converting the equation into intercept form:
\(
\begin{aligned}
& \Rightarrow \frac{x}{\left(\frac{8}{9}\right)}+\frac{y}{\left(-\frac{4}{3}\right)}=1 \\
& \Rightarrow|6 a|+|5 b|=\left|\frac{6 \times 8}{9}\right|+\left|\frac{5 \times(-4)}{3}\right|=12
\end{aligned}
\)
Hence this is the required option.

Hint

\(
y^2=20 x, y=m x+ c
\)
Put value of \(x\)
\(
\begin{aligned}
& y^2=20\left(\frac{y-c}{m}\right) \\
& \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 \ldots \ldots(i)
\end{aligned}
\)
Since, centroid \(=(10,10)\)
So, \(\frac{y_1+y_2+0}{3}=10\)
\(
\Rightarrow y_1+y_2=30
\)
From (i), Sum of roots \(=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3}\)
Also, \(c-m=6 \Rightarrow c=6+\frac{2}{3}=\frac{20}{3}\)
Now, the equation is :
\(
\begin{aligned}
& y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\
& \Rightarrow y^2-30 y+200=0 \\
& \Rightarrow y^2-20 y-10 y+200=0 \\
& \Rightarrow(y-20)(y-10)=0 \\
& \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\
& \therefore P \equiv(5,10), Q \equiv(20,20) \\
& \text { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\
& =225+100=325
\end{aligned}
\)

Hint

Given, Focus \((3,0)\), Directrix, \(x=-3\)
So, the equation of parabola will be, \(y^2=12 x\)
Now taking the parametric point \(\left(3 t^2, 6 t\right)\)
Now we know that \(R(\alpha, \beta)\) will be the aithematic mean of ordinate and geometric mean of abscissa will be,
Hence, \(\beta=\frac{6\left(t_1+t_2\right)}{2}\) and \(\alpha=\sqrt{3 t_1{ }^2 \times 3 t_2{ }^2}=3 t_1 t_2\)
So, putting the value in \(\frac{\beta^2}{\alpha}, \Rightarrow \frac{\beta^2}{\alpha}=\frac{9\left(t_1+t_2\right)^2}{3 t_1 t_2}=\frac{9\left(4 t_2\right)^2}{3 \times 3 t_2{ }^2}=16\left(\because \frac{6 t_1}{6 t_2}=3\right.\), given \()\)

Hint

Given, Common tangent to the curves \(y^2=4 x\) and \((x-4)^2+y^2=16\) touch the curves at the points \(P\) and \(Q\) Let, the equation of tangent be, \(y=m x+\frac{a}{m}\), where \(a=1\) is focus and \(m\) is slope,
So, equation of tangent to parabola will be, \(y=m x+\frac{1}{m}\)
Also line is tangent to circle, So, using perpendicular distance formula from centre \((4,0)\) to tangent which will be radius \(r=4\), we get,
\(
\begin{aligned}
& \left|\frac{4 m+\frac{1}{m}}{m^2+1}\right|=4 \\
& \Rightarrow 16 m^2+\frac{1}{m^2}+8=16 m^2+16 \\
& \Rightarrow m^2=\frac{1}{8} \\
& \Rightarrow m= \pm \frac{1}{2 \sqrt{2}}
\end{aligned}
\)
Now if \(m=\frac{1}{2 \sqrt{2}}\) then \(P=\left(\frac{a}{m^2}, \frac{2 a}{m}\right)=(8,4 \sqrt{2})\)
Now equation of tangent will be, \(x-2 \sqrt{2} y+8=0\)
And \(Q\) will be foot of perpendicular from \((4,0)\) on tangent \(x-2 \sqrt{2} y+8=0\),
So, using foot of perpendicular formula we get,
\(
\begin{aligned}
& \frac{x-4}{1}=\frac{y-0}{-2 \sqrt{2}}=-\frac{4-0+8}{1^2+(2 \sqrt{2})^2} \\
& \Rightarrow(x, y)=\left(\frac{8}{3}, \frac{8 \sqrt{2}}{3}\right)
\end{aligned}
\)
Hence, by distance formula we get,
\(
\begin{aligned}
& (P Q)^2=\left(\frac{16}{3}\right)^2+\left(\frac{4 \sqrt{2}}{3}\right)^2 \\
& \Rightarrow(P Q)^2=\frac{256+32}{9}=32
\end{aligned}
\)

Hint

We have been given the length of focal chord \(P Q=100\).
Length of focal chord at \((t)=a\left(t+\frac{1}{t}\right)^2=100\)
Where \(a=9\)
\(
\begin{aligned}
& t+\frac{1}{t}= \pm \frac{10}{3} \\
& \therefore t=3, \frac{1}{3},-3, \frac{-1}{3}
\end{aligned}
\)
Since ordinate of \(P\) is + ve
\(
\therefore t=3 \text { or } \frac{1}{3}
\)
\(
\Rightarrow P(81,54) \& Q(1,-6)
\)
( \(t=3\) as slope of \(P Q\) is positive)
\(
\therefore M(21,9)
\)
Required line: \(y-9=\frac{-80}{60}(x-21)\)
\(
\Rightarrow 4 x+3 y=111
\)
\((-3,43)\) does not lie on line.
Hence this is the correct option.

Hint

\(
\begin{aligned}
& k=r \\
& h=1
\end{aligned}
\)
Circle touches xaxis at \((1,0)\)
\(\Rightarrow\) radius \(=r=k \Rightarrow\) perpendicular distance from \((1, r)\) to the line \(x+y=0\) is \(\frac{r+1}{\sqrt{2}}\)
\(
OP = r , PR =1
\)
\(
\Rightarrow OR =\frac{ r +1}{\sqrt{2}}
\)
\(
\begin{gathered}
\Rightarrow r^2=1+\frac{(r+1)^2}{2} \\
2 r^2=2+r^2+1+2 r \\
r^2-2 r-3=0 \\
(r-3)(r+1)=0 \\
r=3,-1
\end{gathered}
\)
\(
\text { The value of } h+k+r \text { is equal to } 1+3+3=7 \text {. }
\)

Hint

\(
\begin{aligned}
& L _1: 4 x -3 y + K _1=0 \\
& L _2: 4 x -3 y + K _2=0 \\
& \text { now } \\
& -4-6+ K _1=0 \Rightarrow K _1=10 \\
& 12+18+ K _2=0 \Rightarrow K _2=-30
\end{aligned}
\)
\(\Rightarrow\) Tangent to the circle are
\(
\begin{aligned}
& 4 x-3 y+10=0 \\
& 4 x-3 y-30=0
\end{aligned}
\)
Length of diameter \(2 r=\frac{|10+30|}{5}=8\)
\(
\Rightarrow r =4
\)
Now centre is mid point of \(A \& B\)
\(
x =1, y =-2
\)
Equation of circle
\(
(x-1)^2+(y+2)^2=16
\)

Hint

Equation of the circle with \(PQ\) as diameter is
\(
2\left(x^2+y^2\right)-r x-2 s y+p-2 q=0
\)
on comparing with the given equation
\(
\begin{aligned}
& r =11, s =7 \\
& p -2 q =-22 \\
& \therefore 2 r + s -2 q + p =22+7-22=7
\end{aligned}
\)

Hint

Let \((h, k)\) is centre of circle
\(
\begin{aligned}
& \left|\frac{h-k}{\sqrt{2}}\right|=|h| \\
& k^2-h^2+2 h k=0
\end{aligned}
\)
\(\therefore\) Equation of locus is \(y^2-x^2+2 x y=0\)

Hint

\(
\begin{aligned}
& AB =\sqrt{ } 26 \\
& r ^2= CM ^2+ AM ^2 \\
& =\left(2 \times \sqrt{\frac{13}{2}}\right)^2+\left(\sqrt{\frac{13}{2}}\right)^2 \\
& r ^2=\frac{65}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { C }: 4 x^2+4 y^2-12 x+8 y+k=0 \\
& \Rightarrow x^2+y^2-3 x+2 y+\left(\frac{k}{4}\right)=0 \\
& \text { Centre }\left(\frac{3}{2},-1\right) ; r=\sqrt{\frac{13-k}{2}} \Rightarrow k \leq 13 \ldots (1)
\end{aligned}
\)
(i) Point \(\left(1, \frac{-1}{3}\right)\) lies on or inside circle \(C\)
\(
\Rightarrow S _1 \leq 0 \Rightarrow k \leq \frac{92}{9}
\)
(ii) \(C\) lies in \(4^{\text {th }}\) quadrant

\(
\begin{aligned}
& r<1 \\
& \Rightarrow \frac{\sqrt{13-k}}{2}<1 \\
& \Rightarrow k<9 \ldots .(3)
\end{aligned}
\)
Hence \((1) \cap(2) \cap(3) \Rightarrow k \in\left(9, \frac{92}{9}\right]\)

Hint

\(
\begin{aligned}
& 4 x+3 y+2=0 \\
& 3 x-4 y-11=0
\end{aligned}
\)
\(
\begin{aligned}
& \frac{x}{-25}=\frac{y}{50}=\frac{1}{-25} \\
& \frac{x-1}{\cos \theta}=\frac{y+2}{\sin \theta}= \pm 5 \\
& y=-2+5\left(-\frac{4}{5}\right)=-6 \\
& x=1+5\left(\frac{3}{5}\right)=4
\end{aligned}
\)
Req. distance
\(
\begin{aligned}
& \left|\frac{5(4)-12(-6)+51}{13}\right| \\
& =\left|\frac{20+72+51}{13}\right| \\
& =\frac{143}{13}=11
\end{aligned}
\)

Hint

Tangent at \(O\)
\(
\begin{aligned}
& -(x+0)-2(y+0)=0 \\
& \Rightarrow x+2 y=0
\end{aligned}
\)
Tangent at \(P\)
\(
x(1+\sqrt{5})+y \cdot 2-(x+1+\sqrt{5})-2(y+2=0)
\)
Put \(x=-2 y\)
\(
\begin{aligned}
& -2 y(1+\sqrt{5})+2 y+2 y-1-\sqrt{5}-2 y-4=0 \\
& -2 \sqrt{5} y=5+\sqrt{5} \Rightarrow y=\left(\frac{\sqrt{5}+1}{2}\right)
\end{aligned}
\)
\(
Q\left(\sqrt{5}+1,-\frac{\sqrt{5}+1}{2}\right)
\)
Length of tangent \(O Q=\frac{5+\sqrt{5}}{2}\)
\(
\begin{aligned}
& \text { Area }=\frac{ RL ^3}{ R ^2+ L ^2} \\
& R =\sqrt{5} \\
& =\frac{\sqrt{5} \times\left(\frac{5+\sqrt{5}}{2}\right)^3}{5+\left(\frac{5+\sqrt{5}}{2}\right)^2} \\
& =\frac{\sqrt{5}}{2} \times \frac{4 \times(125+75+75 \sqrt{5}+5 \sqrt{5})}{(20+25+10 \sqrt{5}+5)} \\
& =\frac{5+3 \sqrt{5}}{2}
\end{aligned}
\)

Hint

Normal are
\(
\begin{aligned}
& y+2 x=\sqrt{11}+7 \sqrt{7} \\
& 2 y+x=2 \sqrt{11}+6 \sqrt{7}
\end{aligned}
\)
Center of the circle is point of intersection of normals i.e.
\(
\left(\frac{8 \sqrt{7}}{3}, \sqrt{11}+\frac{5 \sqrt{7}}{3}\right)
\)
Tangent is \(\sqrt{11} y-3 x=\frac{5 \sqrt{77}}{3}+11\)
Radius will be \(\perp\) distance of tangent from center i.e. \(4 \sqrt{\frac{7}{5}}\)
Now \((5 h-8 k)^2+5 r^2=816\)

Hint

\(PQ\) is diameter of circle
\(
\begin{aligned}
& S: x^2+y^2-2 \sqrt{2} x-6 \sqrt{2} y+14=0 \\
& C(\sqrt{2}, 3 \sqrt{2}), O(2 \sqrt{2}, 2 \sqrt{2}) \\
& r_1=\sqrt{6} \\
& S_1:(x-2 \sqrt{2})^2+(y-2 \sqrt{2})^2=r^2
\end{aligned}
\)
Now in \(\triangle OCQ\)
\(
\begin{aligned}
& | OC |^2+| CQ |^2=| OQ |^2 \\
& 4+6= r ^2 \\
& r ^2=10
\end{aligned}
\)

Hint

\(
OP =\left|\frac{2-3+2}{\sqrt{2}}\right|
\)

\(
\begin{aligned}
& OP =\frac{3}{\sqrt{2}} \\
& AP =\sqrt{ OA ^2- OP ^2} \\
& =\frac{1}{\sqrt{2}} \\
& \tan \theta=3 \\
& \therefore \sin \theta=\frac{3}{\sqrt{10}}=\frac{ AP }{ AN } \\
& \Rightarrow AN =\frac{\sqrt{5}}{3}= BN \\
& \text { Area of } \triangle ANB =\frac{1}{2} \cdot\left( AN ^2\right) \sin 2 \theta=\frac{1}{6}
\end{aligned}
\)

Hint

Radius of given circle is 1 .
\(
\begin{aligned}
& BC =\text { diameter }=2 . AB =\sqrt{2} \\
& AC =\sqrt{ BC ^2- AB ^2}=\sqrt{2} \\
& \triangle ABC =\frac{1}{2} AB \cdot AC =1
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A =\frac{1}{2} a (1-\cos \theta)(4 \sin \theta) \\
& A =2 a (1-\cos \theta) \sin \theta \\
& \frac{ dA }{ d \theta}=2 a \left(\sin ^2 \theta+\cos \theta-\cos ^2 \theta\right) \\
& \frac{ dA }{ d \theta}=0 \Rightarrow 1+\cos \theta-2 \cos ^2 \theta=0 \\
& \cos \theta=1 \text { (Reject) }
\end{aligned}
\)
or
\(
\begin{aligned}
& \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \\
& \frac{ d ^2 A }{ d \theta^2}=2 a \left(2 \sin ^2 \theta-\sin \theta\right) \\
& \frac{ d ^2 A }{ d \theta^2}<0 \text { for } \theta=\frac{2 \pi}{3}
\end{aligned}
\)
Now, \(A_{\max }=\frac{3 \sqrt{3}}{2} a=6 \sqrt{3}\)
\(
a=4
\)
Now, \(e=\sqrt{\frac{a^2-b^2}{a^2}}=\frac{\sqrt{3}}{2}\)

Hint

Ellipse \(x^2+2 y^2=4\)
Line \(y=x+1\)
Point of intersection
\(
\begin{aligned}
& x^2+2(x+1)^2=4 \\
& 3 x^2+4 x-2=0 \\
& \left|x_1-x_2\right|=\frac{\sqrt{40}}{3} \\
& A B=2 r=\left|x_1-x_2\right| \sqrt{1+m^2}
\end{aligned}
\)
\(m\) is slope of given line
\(
\begin{aligned}
& AB =\frac{\sqrt{40}}{3} \sqrt{1+1} \\
& 2 r =\frac{\sqrt{80}}{3} \Rightarrow r =\frac{\sqrt{80}}{6} \\
& (3 r )^2=\left(3 \times \frac{\sqrt{80}}{6}\right)^2=\frac{80}{4}=20
\end{aligned}
\)