Ellipse

$ \begin{aligned} &\text { Equation of ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ &\begin{aligned} & \text { At }(2,2) \Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1 \\ & \Rightarrow\left(1-\frac{4}{b^2}\right) \frac{1}{4}=\frac{1}{a^2} \\ & \text { At }(3,1) \Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1 \\ & \Rightarrow \frac{9}{4}\left(1-\frac{4}{b^2}\right)+\frac{1}{b^2}=1 \\ & \Rightarrow \quad \frac{5}{4}=\frac{8}{b^2} \Rightarrow b^2=\frac{32}{5} \\ & \text { and } a^2=\frac{32}{3} \\ & \Rightarrow 3 a^2+5 b^2=32+32=64 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } y=4 x+c \text { touches ellipse }\\ &\begin{aligned} \frac{x^2}{4}+\frac{y^2}{1} & =1 \\ c & = \pm \sqrt{a^2 m^2+b^2} \\ c & = \pm \sqrt{4(4)^2+1}= \pm \sqrt{65} \end{aligned} \end{aligned} $

If, $x \cos \alpha+y \sin \alpha=p$ is the tangent of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then

$ p^2=a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha $

∴ From the given question

$ \begin{aligned} p & =2 \sqrt{3}, a^2=16, b^2=8 \\ \therefore \quad(2 \sqrt{3})^2 & =16 \cos ^2 \alpha+8 \sin ^2 \alpha \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 12=8 \cos ^2 \alpha+8 \\ & \Rightarrow 8 \cos ^2 \alpha=4 \Rightarrow \cos ^2 \alpha=\frac{1}{2} \\ & \Rightarrow \quad \cos \alpha= \pm \frac{1}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} &\because \text { a is an acute angle. }\\ &\begin{aligned} & a=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ & a=\frac{\pi}{4} \end{aligned} \end{aligned} $

$ x+\sqrt{3} y=3 $

$ \begin{aligned} & \Rightarrow y=\frac{-1}{\sqrt{3}} x+\sqrt{3} \\ & \therefore \text { Slope of tangent }=-\frac{1}{\sqrt{3}} \end{aligned} $

from the equation of ellipse,

$ \begin{array}{rlrl} & 4 x+6 y\left(\frac{d y}{d x}\right)= 0 \\ \Rightarrow & \frac{d y}{d x}= \frac{-2 x}{3 y}= \text { Slope of tangent } \\ & -\frac{1}{\sqrt{3}}= \frac{-2 x}{3 y} \\ & \sqrt{3} y= 2 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Which will satisfy the equation of angent.

$ \begin{aligned} \therefore & & x+2 x & =3 \\ \Rightarrow & & x & =1 \end{aligned} $

Then, $y=\frac{2}{\sqrt{3}}$

So, coordinates of $p=\left(1, \frac{2}{\sqrt{3}}\right)$

$ \begin{aligned} \text { and slope of normal } & =\frac{-1}{\text { Slope of tangent }} \\ & =\sqrt{3} \end{aligned} $

Then, equation of normal,

$ y=\sqrt{3} x+C $

Which is going through point $P$.

$ \begin{array}{ll} \therefore & \frac{2}{\sqrt{3}}=\sqrt{3}+C \\ \Rightarrow & C=\frac{-1}{\sqrt{3}} \end{array} $

Therefore, equation of normal,

$ y=\sqrt{3} x-\frac{1}{\sqrt{3}} \Rightarrow 3 x-\sqrt{3} y=1 $

Given equation

$ S=2 x^2-x y+y^2+2 x+3 y+1=0 $

When shifted $(h, k), x=x+h, y=y+k$, then

$ \begin{gathered} S=2(x+h)^2-(x+h)(y+k)+(y+k)^2 \\ +2(x+h)+3(y+k)+1=0 \\ \Rightarrow 2\left(x^2+2 x h+h^2\right)-(x y+x k+y h+h k) \\ +y^2+2 k y+k^2+2 x+2 h+3 y+3 k+1=0 \\ \Rightarrow 2 x^2+4 x h+2 h^2-x y-x k-y h-h k+y^2 \\ +2 k y+k^2+2 x+2 h+3 y+3 k+1=0 \\ \Rightarrow 2 x^2-x y+y^2+(4 h-k+2 x \\ +(-h+2 k+3) y+2 h^2-h k+k^2 \\ +2 h+3 k+1=0 \end{gathered} $

This equation equal to

$ a x^2+2 h x y+b y^2-3=0 $

$ \begin{array}{r}So, 4 k-k+2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -h+2 k+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

On solving Eqs. (i) and (ii), we get

$ h=-1, k=-2, b=2, a=4 $

Now, $4 x^2-2 x y+2 y^2-3=0$ is rotated with angle $\theta$.

Replace $x$ by $x \cos \theta-y \sin \theta$ and $y$ by $x \sin \theta+y \cos \theta$

$ \begin{gathered} 4(x \cos \theta-y \sin \theta)^2-2(x \cos \theta-y \sin \theta) \\ (x \sin \theta+y \cos \theta) \\ +2(x \cos \theta+y \sin \theta)^2-3=0 \\ \Rightarrow 4\left(x^2 \cos ^2 \theta+y^2 \sin ^2 \theta-2 y \sin \theta \cos \theta\right) \\ -2\left(x^2 \sin \theta \cos \theta-y^2 \sin \theta \cos \theta\right) \end{gathered} $

$ \begin{array}{r} +x y\left(\cos ^2 \theta-\sin ^2 \theta\right)+2\left(x^2 \cos ^2 \theta+y^2 \sin ^2 \theta\right. \\ +2 x y \sin \theta \cos \theta)-3=0 \\ \end{array} $

$ \begin{aligned} & \Rightarrow\left(4 \cos ^2 \theta-2 \sin \theta \cos \theta+2 \cos ^2 \theta\right) x^2 \\ & \quad+\left(4 \sin ^2 \theta-2 \sin \theta \cos \theta+2 \sin ^2 \theta\right) \end{aligned} $

$ \begin{array}{r} -8 x y \sin \theta \cos \theta-2 x y \cos ^2 \theta \sin ^2 \theta \\ +4 x y \sin \theta \cos \theta-3=0 \\ \Rightarrow-8 \sin \theta \cos \theta-2\left(\cos ^2 \theta-\sin ^2 \theta\right) \\ +4 \sin \theta \cos \theta=0 \\ \Rightarrow-2 \sin 2 \theta-2 \cos 2 \theta=0 \Rightarrow \tan 2 \theta=-1 \\ \therefore h+k+\tan 2 \theta=-1-2-1=-4 \end{array} $

Consider the equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a>b$.

Given, $A S=4-\sqrt{7}$

$ \begin{aligned} \Rightarrow \quad a-a c & =4-\sqrt{7} \\ B S & =4 \Rightarrow B S^{\prime}=4 \end{aligned} $

According to the definition of ellipse

$ \begin{array}{ll} & B S^{\prime}+B S=2 a \\ \Rightarrow & 4+4=2 a \Rightarrow a=4 \\ \text { As, } & a=4 \\ \Rightarrow & 4-4 e=4-\sqrt{7} \Rightarrow e=\sqrt{7} / 4 \\ \Rightarrow & a e=\sqrt{7} \end{array} $

Consider the $\triangle B S S^{\prime}$

Now, by using the cosine formula, we get

$ \begin{aligned} \cos \theta & =\frac{4^2+4^2-(2 \sqrt{7})^2}{2(4)(4)} \\ & =\frac{32-28}{8(4)}=\frac{4}{8(4)}=\frac{1}{8} \end{aligned} $

Given, $x=1+2 \cos \theta$

$ \begin{aligned} & \Rightarrow \quad\left(\frac{x-1}{2}\right)^2=\cos ^2 \theta \text { and } y=2+\sin \theta \\ & \Rightarrow \quad(y-2)^2=\sin ^2 \theta \end{aligned} $

Thus, the described equation of the ellipse be

$ \frac{(x-1)^2}{4}+\frac{(y-2)^2}{1}=1 $

The major axis of this ellipse is $y=2$.

Now, $P\left(\frac{\pi}{4}\right)=\left(1+2 \cos \frac{\pi}{4}, 2+\sin \frac{\pi}{4}\right)$

$ =\left(1+\frac{2}{\sqrt{2}}, 2+\frac{1}{\sqrt{2}}\right) $

The equation of the normal at $\left(1+\frac{2}{\sqrt{2}}, 2+\frac{1}{\sqrt{2}}\right)$ to this ellipse be

$ \frac{(4) x}{1+\frac{2}{\sqrt{2}}}-\frac{(\text { l) } y}{2+\frac{1}{\sqrt{2}}}=4-1 $

On putting $y=2$, we get

$ \begin{aligned} & \frac{4 \sqrt{2} x}{2+\sqrt{2}}-\frac{2 \sqrt{2}}{2 \sqrt{2}+1}=3 \\ & x=\left(3+\frac{2 \sqrt{2}}{2 \sqrt{2}+1}\right)\left(\frac{2+\sqrt{2}}{4 \sqrt{2}}\right) \\ & x=2.26 \end{aligned} $

Thus, the intersection point be $(2.26,2)$

Given, points $A=(2,0)$

and $B=(0,-2)$

Let $P(x, y)$ be any variable point, then

$ \begin{aligned} & P A+P B=4 \\ & \Rightarrow \sqrt{(x-2)^2+y^2}+\sqrt{x^2+(y+2)^2}=4 \\ & \Rightarrow \sqrt{(x-2)^2+y^2}=4-\sqrt{x^2+(y+2)^2} \\ & \Rightarrow(x-2)^2+y^2=16+x^2+(y+2)^2 -8 \sqrt{x^2+(y+2)^2} \\ & \Rightarrow x^2+4-4 x+y^2=16+x^2+y^2 +4 y+4-8 \sqrt{x^2+y^2+4 y+4} \end{aligned} $

$ \begin{array}{lr} \Rightarrow & 16+4 y+4 x=8 \sqrt{x^2+y^2+4 y+4} \\ \Rightarrow & 4+x+y=2 \sqrt{x^2+y^2+4 y+4} \\ \Rightarrow & 16+x^2+y^2+8 x+2 x y+8 y \\ & =4 x^2+4 y^2+16 y+16 \\ \Rightarrow & 3 x^2-2 x y+3 y^2-8 x+8 y=0 \end{array} $

To shift origin so that first degree lerms are eliminated.

substitute $x$ with $x+h$ and $y$ with $y+k$

$ \begin{gathered} +4(y+k)+4=0 \\ 3(x+h)^2+(y+k)^2-6(x+h) \\ 3\left(x^2+h^2+2 x h\right)+\left(y^2+2 k y+k^2\right) \\ -6(x+h)+4(y+k)+4=0 \end{gathered} $

Equating coefficients of first degree term to zero

$ \begin{aligned} 6 h-6 & =0 \text { and } 2 k+4=0 \\ \Rightarrow h=1 \text { and } k & =-2 \end{aligned} $

∴ Origin is shifted to $P$ in

$ \begin{aligned} & 2 x^2+3 x y-5 y^2+2 x-23 y-24=0 \\ & \Rightarrow 2(x+1)^2+3(x+1)(y-2)-5(y-2)^2 \\ & \quad+2(x+1)-23(y-2)-24=0 \\ & \Rightarrow 2\left(x^2+2 x+1\right)+3(x y-2 x+y-2)-5 \\ & \left(y^2-4 y+4\right)+2 x+2-23 y+46-24=0 \\ & \Rightarrow 2 x^2+3 x y-5 y^2=0 \end{aligned} $

$ \begin{aligned} &\text { Let the equation of ellipse be }\\ &\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{aligned} $

$ \begin{aligned} O S & =a e, O B=b \Rightarrow \angle S^{\prime} S B=\frac{\pi}{6} \\ \tan \frac{\pi}{6} & =\frac{O B}{O S}=\frac{b}{a e} \Rightarrow \frac{1}{\sqrt{3}}=\frac{b}{a e} \\ \Rightarrow \quad e \quad e & =\frac{b}{a} \sqrt{3} \end{aligned} $

Also, $e^2=1-\frac{b^2}{a^2} \Rightarrow e^2=1-\frac{e^2}{3}$

$ \begin{aligned} & \Rightarrow \quad \frac{4}{3} e^2=1 \Rightarrow e^2=\frac{3}{4} \\ & \Rightarrow \frac{a^2-b^2}{a^2}=\frac{3}{4} \Rightarrow a^2=4 b^2 \end{aligned} $

Also, ellipse passes through $(2 \sqrt{3}, 1)$.

$ \begin{aligned} & \therefore \quad \frac{12}{a^2}+\frac{1}{b^2}=1 \Rightarrow 12 b^2+a^2=a^2 b^2 \\ & \Rightarrow \quad 12 b^2+4 b^2=4 b^4 \Rightarrow 16 b^2=4 b^4 \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & b^2=4, b \neq 0 \Rightarrow b= \pm 2 \\ \therefore & a^2=16 \\ \text { or } & a= \pm 4 \end{array}\\ &\text { Sum of squares of major axis and minor }\\ &\text { axis }=8^2+4^2=64+16=80 \end{aligned} $

Equation of normal to the ellipse

$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { is } y=m x+C $

If $c^2=\frac{m^2\left(a^2-b^2\right)^2}{a^2+b^2 m^2}$

Here, given ellipse

$ \frac{x^2}{36}+\frac{y^2}{16}=1 $

and normal is

$ 4 x+2 y+n=0 $

or  $ y=-2 x-\frac{n}{2} $

$ \begin{aligned} &\text { On comparing, we have }\\ &\begin{aligned} & & a^2 & =36, b^2=16, m=-2, c=-\frac{n}{2} \\ \therefore & & \frac{n^2}{4} & =\frac{4(36-16)^2}{36+16(-2)^2} \Rightarrow \frac{n^2}{4}=4 \times 2 \times 2 \\ \Rightarrow & & n & = \pm 8 \end{aligned} \end{aligned} $

Given equation of ellipse is

$ \begin{aligned} & x^2+2 y^2=2 \\ \Rightarrow \quad & \frac{x^2}{2}+\frac{y^2}{1}=1 \end{aligned} $

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

This meets the axis at the points $A(a \sec \theta, 0)$ and $B(0, b \operatorname{cosec} \theta)$.

Now, Let $\left(x_1, y_1\right) x$ mid-point of $A B$.

Therefore, $2 x_1=a \sec \theta$ and

$ 2 y_1=b \operatorname{cosec} \theta $

$ \begin{aligned} &\begin{aligned} & \Rightarrow 2 \cos \theta=\frac{a}{x_1} \text { and } 2 \sin \theta=\frac{b}{y_1} \\ & \Rightarrow \quad 4\left(\cos ^2 \theta+\sin ^2 \theta\right)=\frac{a^2}{x_1^2}+\frac{b^2}{y_1^2} \\ & \Rightarrow \quad 4=\frac{a^2}{x_1^2}+\frac{b^2}{y_1^2} \Rightarrow 4=\frac{2}{x_1^2}+\frac{1}{y_1^2} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(\because a^2=2, b^2=1\right) \end{aligned}\\ &\text { Therefore, locus of }\left(x_1, x_2\right) \text { is }\\ &4=\frac{2}{x^2}+\frac{1}{y^2} \Rightarrow \frac{1}{2 x^2}+\frac{1}{4 y^2}=1 \end{aligned} $

The equation of ellipse is

$ \frac{x^2}{9}+\frac{y^2}{25}=1 $

Now, equation of tangent at any point ( $3 \cos \theta, 5 \sin \theta$ ) on the ellipse is

$ \frac{x}{3} \cos \theta+\frac{y}{5} \sin \theta=1 $

i.e. $5 x \cos \theta+3 y \sin \theta-15=0$

The product of perpendicular from $S(3 e, 0)$ and $S(-3 e, 0)$ on this tangent is

$ \begin{aligned} & \frac{(15 e \cos \theta-15)(-15 e \cos \theta-15)}{25 \cos ^2 \theta+9 \sin ^2 \theta} \\ & \quad=\frac{225\left(1-e^2 \cos ^2 \theta\right)}{25 \cos ^2 \theta+25\left(1-e^2\right) \cos ^2 \theta} \\ & \quad=\frac{225\left(1-e^2 \cos ^2 \theta\right)}{25\left(1-e^2 \cos ^2 \theta\right)}=9 \end{aligned} $

The equation of ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$

So, $e=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

So, $a e=2$

Now, equation of tangent at $\left(2, \frac{5}{3}\right)$ is

$ T:\left(\frac{2}{9}\right) x+\frac{y}{3}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and equation of tangent at $\left(2,-\frac{5}{3}\right)$ is

$ T^1:\left(\frac{2}{9}\right) x+\left(-\frac{y}{3}\right)=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i),

$x$ intercept, $y=0 \Rightarrow x=\frac{9}{2}$

$y$ intercept, $x=0 \Rightarrow y=3$

Now, area of $\triangle A O B$ is

$ =\frac{1}{2} \times O B \times O A=\frac{1}{2} \times \frac{9}{2} \times 3=\frac{27}{4} $

Thus, area of quadrilateral

$ \begin{aligned} & =4 \times(\text { area of } \triangle A O B) \\ & =4 \times \frac{27}{4}=27 \end{aligned} $

The particle is moving clockwise along the ellipse described by the equation:

$ \frac{x^2}{100} + \frac{y^2}{25} = 1 $

When the particle reaches the point $(-8, 3)$ on the ellipse, it moves along the tangent line at this location. We need to determine where this tangent line intersects the $Y$-axis.

Equation of the Tangent Line:

For an ellipse with the general form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of the tangent line at any point $(x_1, y_1)$ on the ellipse is given by:

$ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 $

Substituting $a^2 = 100$, $b^2 = 25$, and the point $(-8, 3)$, the equation of the tangent becomes:

$ \frac{x(-8)}{100} + \frac{y(3)}{25} = 1 $

Simplifying:

$ -2x + 3y = 25 $

Finding the Intersection with the $Y$-Axis:

To find where the tangent line crosses the $Y$-axis, we set $x = 0$:

$ -2(0) + 3y = 25 \Rightarrow 3y = 25 \Rightarrow y = \frac{25}{3} $

Thus, the particle crosses the $Y$-axis at the point:

$ \left(0, \frac{25}{3}\right) $

$P F_1+P F_2=2 a$, where $a$ is length of semi-major axis.

$ \begin{aligned} & \because \text { Ellipse passes through }(0,0) \\ & \begin{array}{l} \Rightarrow \sqrt{(-4)^2+4^2}+\sqrt{3^2+3^2}=2 a \\ \Rightarrow 4 \sqrt{2}+3 \sqrt{2}=2 a \Rightarrow a=\frac{7 \sqrt{2}}{2}=\frac{7}{\sqrt{2}} \\ \qquad F_1 F_2=2 a e \end{array} \\ & \Rightarrow \sqrt{(-4-3)^2+(4-3)^2}=2 \times \frac{7}{\sqrt{2}} \times e \\ & \Rightarrow \sqrt{50}=\sqrt{2} \times 7 e \Rightarrow e=\frac{5}{7} \end{aligned} $

${L_1}:bx + 10y - 8 = 0,\,{L_2}:2x - 3y = 0$

then $L:(bx + 10y - 8) + \lambda (2x - 3y) = 0$

$\because$ It passes through $(1,\,1)$

$\therefore$ $b + 2 - \lambda = 0 \Rightarrow \lambda = b + 2$

and touches the circle ${x^2} + {y^2} = {{16} \over {17}}$

$\left| {{{{8^2}} \over {{{(2\lambda + b)}^2} + {{(10 - 3\lambda )}^2}}}} \right| = {{16} \over {17}}$

$ \Rightarrow 4{\lambda ^2} + {b^2} + 4b\lambda + 100 + 9{\lambda ^2} - 60\lambda = 68$

$ \Rightarrow 13{(b + 2)^2} + {b^2} + 4b(b + 2) - 60(b + 2) + 32 = 0$

$ \Rightarrow 18{b^2} = 36$

$\therefore$ ${b^2} = 2$

$\therefore$ Eccentricity of ellipse : ${{{x^2}} \over 5} + {{{y^2}} \over {{b^2}}} = 1$ is

$\therefore$ $e = \sqrt {1 - {2 \over 5}} = \sqrt {{3 \over 5}} $

$2{x^2} + 3{y^2} = 5$

Equation of tangent having slope m.

$y = mx\, \pm \,\sqrt {{5 \over 2}{m^2} + {5 \over 3}} $

which passes through $(1,3)$

$3 = m\, \pm \sqrt {{5 \over 2}{m^2} + {5 \over 3}} $

${5 \over 2}{m^2} + {5 \over 3} = 9 + {m^2} - 6m$

${3 \over 2}{m^2} + 6m - {{22} \over 3} = 0$

$9{m^2} + 36m - 44 = 0$

${m_1} + {m_2} = - 4,\,{m_1}{m_2} = - {{44} \over 9}$

${({m_1} - {m_2})^2} = 16 + 4 \times {{44} \over 9} = {{320} \over 9}$

Acute angle between the tangents is given by

$\alpha = {\tan ^{ - 1}}\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$

$ = {\tan ^{ - 1}}\left| {{{{{8\sqrt 5 } \over 3}} \over {1 - {{44} \over 9}}}} \right|$

$ = {\tan ^{ - 1}}\left( {{{24\sqrt 5 } \over {35}}} \right)$

$\alpha = {\tan ^{ - 1}}\left( {{{24} \over {7\sqrt 5 }}} \right)$

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ meets the line ${x \over 7} + {y \over {2\sqrt 6 }} = 1$ on the x-axis

So, $a = 7$

and ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ meets the line ${x \over 7} - {y \over {2\sqrt 6 }} = 1$ on the y-axis

So, $b = 2\sqrt 6 $

Therefore, ${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {{24} \over {49}}$

$e = {5 \over 7}$

Given ellipse ${x^2} + {a^2}{y^2} = 25{a^2} \Rightarrow {{{x^2}} \over {25{a^2}}} + {{{y^2}} \over {25}} = 1$

eccentricity $({e_1}) = \sqrt {1 - {{{b^2}} \over {{a^2}}}} $

$ = \sqrt {1 - {{25} \over {25{a^2}}}} $

$ = \sqrt {1 - {1 \over {{a^2}}}} $

$ \Rightarrow e_1^2 = 1 - {1 \over {{a^2}}}$

Also, given hyperbola,

${x^2} - {a^2}{y^2} = 5$

$ \Rightarrow {{{x^2}} \over 5} - {{{y^2}} \over {{5 \over {{a^2}}}}} = 1$

eccentricity $({e_2}) = \sqrt {1 + {{{b^2}} \over {{a^2}}}} $

$ = \sqrt {1 + {5 \over {5{a^2}}}} $

$ = \sqrt {1 + {1 \over {{a^2}}}} $

$ \Rightarrow e_2^2 = 1 + {1 \over {{a^2}}}$

Also given,

${e_1} = b \times {e_2}$

$ \Rightarrow e_1^2 = {b^2} \times e_2^2$

$ \Rightarrow 1 - {1 \over {{a^2}}} = {b^2}\left( {1 + {1 \over {{a^2}}}} \right)$

$ \Rightarrow {{{a^2} - 1} \over {{a^2}}} = {{{b^2}({a^2} + 1)} \over {{a^2}}}$

$ \Rightarrow {b^2} = {{{a^2} - 1} \over {{a^2} + 1}}$

$y = \log _e^x$ is inverse of $y = {e^x}$ so it is mirror image of each other with respect to y = x line.

Slope of tangent to y = e^x curve

${{dy} \over {dx}} = {e^x}$

Slope of tangent to $y = \log _e^x$ curve,

${{dy} \over {dx}} = {1 \over x}$

Both tangents are parallel to y = x line for minimum distance condition.

$\therefore$ Slope of y = x line = Slope of both the tangent.

$\therefore$ ${{dy} \over {dx}} = {e^x} = 1 \Rightarrow {e^x} = {e^0} = x = 0$

$\therefore$ $y = {e^x} = {e^0} = 1$

and ${{dy} \over {dx}} = {1 \over x} = 1 \Rightarrow x = 1$

$\therefore$ $y = \log _e^1 = 0$

$\therefore$ tangent at (0, 1) point of $y = {e^x}$ curve and tangent at (1, 0) point of $y = \log _e^x$ curve are parallel.

$\therefore$ Minimum distance between point (0, 1) and (1, 0) is $ = \sqrt {{1^2} + {1^2}} = \sqrt 2 $

$\therefore$ $a = \sqrt 2 $

So, ${b^2} = {{{a^2} - 1} \over {{a^2} + 1}} = {{2 - 1} \over {2 + 1}} = {1 \over 3}$

$\therefore$ ${a^2} + {1 \over {{b^2}}} = 2 + {1 \over {1/3}} = 5$

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$

$ \Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1$

$ \Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1$ ..... (i)

${a^2}(1 - {e^2}) = {b^2}$

${a^2}\left( {1 - {1 \over {16}}} \right) = {b^2}$

$15{a^2} = 16{b^2} \Rightarrow {a^2} = {{16{b^2}} \over {15}}$

From (i)

${6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15$ & ${a^2} = 16$

${a^2} + {b^2} = 15 + 16 = 31$

${C_1}:{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$ and ${C_2}:{x^2} + {y^2} = 12$

Let $y = mx \pm \,\sqrt {16{m^2} + 9} $ be any tangent to C₁ and if this is also tangent to C₂ then

$\left| {{{\sqrt {16{m^2} + 9} } \over {\sqrt {{m^2} + 1} }}} \right| = \sqrt {12} $

$ \Rightarrow 16{m^2} + 9 = 12{m^2} + 12$

$ \Rightarrow 4{m^2} = 3 \Rightarrow 12{m^2} = 9$

Let $P(2\cos \theta ,\,\sqrt 2 \sin \theta )$ be any point on ellipse ${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$ and Q(4, 3) and let (h, k) be the mid point of PQ

then $h = {{2\cos \theta + 4} \over 2},\,k = {{\sqrt 2 \sin \theta + 3} \over 2}$

$\therefore$ $\cos \theta = h - 2,\,\sin \theta = {{2k - 3} \over {\sqrt 2 }}$

$\therefore$ ${(h - 2)^2} + {\left( {{{2k - 3} \over {\sqrt 2 }}} \right)^2} = 1$

$ \Rightarrow {{{{(x - 2)}^2}} \over 1} + {{{{\left( {y - {3 \over 2}} \right)}^2}} \over {{1 \over 2}}} = 1$

$\therefore$ $e = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$

Let point (a, a + 1) as the point of intersection of line and ellipse.

So, ${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$

$ \Rightarrow 3{a^2} + 4a - 2 = 0$

If roots of this equation are $\alpha$ and $\beta$.

So, $P(\alpha ,\,\alpha + 1)$ and $Q(\beta ,\,\beta + 1)$

$PQ = 4{r^2} = {(\alpha - \beta )^2} + {(\alpha - \beta )^2}$

$ \Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha - \beta )^2})$

$ = {9 \over 2}\left[ {{{(\alpha + \beta )}^2} - 4\alpha \beta } \right]$

$ = {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$

$ = {1 \over 2}[16 + 24] = 20$

Given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 4} = 1,\,a > 2$

$\therefore$ Let A($\theta$) be the area of $\Delta$ABB'

Then $A(\theta ) = {1 \over 2}4\sin \theta (a + a\cos \theta )$

$A'(\theta ) = a(2\cos \theta + 2{\cos ^2}\theta )$

For maxima $A'(\theta ) = 0$

$ \Rightarrow \cos \theta = 1,\,\,\cos \theta = {1 \over 2}$

But for maximum area $\cos \theta = {1 \over 2}$

$\therefore$ $A(\theta ) = 6\sqrt 3 $

$ \Rightarrow 2{{\sqrt 3 } \over 2}\left( {a + {a \over 2}} \right) = 6\sqrt 3 $

$ \Rightarrow a = 4$

$\therefore$ $e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over {16}}} = {{\sqrt 3 } \over 2}$

Given, equation of ellipse

$ \begin{aligned} & 25 x^2+16 y^2-150 x-64 y-111=0 \\ \Rightarrow & 25 x^2-150 x+225+16 y^2-64 y+64 \\ = & 111+225+64 \\ \Rightarrow \quad & (5 x-15)^2+(4 y-8)^2=400 \\ \Rightarrow \quad & 25(x-3)^2+16(y-2)^2=400 \\ \Rightarrow \quad & \frac{(x-3)^2}{400}+\frac{(y-2)^2}{400}=1 \\ \Rightarrow & \frac{(x-3)^2}{16}+\frac{(y-4)^2}{25}=1 \end{aligned} $

Let $x-3=X$ and $y-2=Y$

$ \therefore \quad \frac{X^2}{16}+\frac{Y^2}{25} $

On comparing with $\frac{X^2}{b^2}+\frac{Y^2}{a^2}=1$

$\therefore a^2=25$ and $b^2=16$

Length of latusrectum $(m)=2 b^2 / a$

$ =\frac{2 \times 16}{5}=\frac{32}{5} $

and length of the major axis $(h)=2 a$

$ \begin{aligned} & =2 \times 5=10 \\ \therefore \quad(m, n) & =\left(\frac{32}{5}, 10\right) \end{aligned} $

Given, equation of ellipse is

$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $

Here, coordinate of $P$ is $(a \cos \theta, b \sin \theta)$ and coordinate of $Q$ is

$ \begin{gathered} \left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right) \\ =(-a \sin \theta, b \cos \theta) \end{gathered} $

Let $R(h, k)$ be the mid-point of $P Q$.

$ \begin{aligned} &\begin{array}{ll} \therefore & h=\frac{a \cos \theta-a \sin \theta}{2}, \\ & k=\frac{b \sin \theta+b \cos \theta}{2} \\ \Rightarrow & \frac{2 h}{a}=\cos \theta-\sin \theta, \\ & \frac{2 k}{b}=\sin \theta+\cos \theta \\ \therefore & \frac{4 h^2}{a^2}+\frac{4 k^2}{b^2}=2 \Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=\frac{1}{2} \end{array}\\ &\text { ∴ Locus of mid-point of } P Q \text { is }\\ &\begin{aligned} \frac{x^2}{a^2}+\frac{y^2}{b^2} & =\frac{1}{2} \\ \Rightarrow \quad \frac{x^2}{a^2 / 2}+\frac{y^2}{b^2 / 2} & =1 \\ \frac{a+b}{\alpha+\beta} & =\frac{a+b}{\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}}=\sqrt{2} \end{aligned} \end{aligned} $

Given, length of latusrectum $=6$

$ \begin{aligned} & & \frac{2 b^2}{a} & =6 \\ \Rightarrow & & \frac{b^2}{a} & =3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Distance between focus and nearest vertex on major-axis $=\frac{5}{3}$

$ \begin{array}{rr} & a-a e=\frac{5}{3} \\ \Rightarrow & a(1-e)=\frac{5}{3}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array} $

$ \begin{aligned} &\begin{array}{lc} & e^2=1-\frac{b^2}{a^2} \\ \Rightarrow & e^2=1-\frac{b^2}{a} \cdot \frac{1}{a} \\ \Rightarrow & e^2=1-3 \times \frac{3}{5}(1-e) \\ \Rightarrow & 5 e^2=5-9+9 e \\ \Rightarrow & 5 e^2-9 e+4=0 \\ \Rightarrow & 5 e^2-5 e-4 e+4=0 \\ \Rightarrow & (5 e-4)(e-1)=0 \\ \Rightarrow & e \neq 1, \text { so, } e=\frac{4}{5} \end{array}\\ &\text { ∴ e satisfies } 25 x^2-40 x+16=0 \end{aligned} $

Given, ellipse

$ \begin{aligned} & x=3 \cos \theta \\ & y=5 \sin \theta \end{aligned} $

Since, $\sin ^2 \theta+\cos ^2 \theta=1$

$ \begin{array}{lc} \Rightarrow & \frac{x^2}{3^2}+\frac{y^2}{5^2}=1 \\ \Rightarrow & 25 x^2+9 y^2=225 &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Line $2 x-3 y+4=0$ cuts ellipse in $A$ and $B$

So, substituting $y=\frac{2 x+4}{3}$ in Eq. (i), we get

$ \begin{array}{rlrl} 25 x^2+9\left(\frac{2 x+4}{3}\right)^2 & =225 \\ \Rightarrow 25 x^2+4 x^2+16+16 x & =225 \\ \Rightarrow & 29 x^2+16 x-209 & =0 \\ \Rightarrow & x_1+x_2 =-\frac{16}{29} \end{array} $

Similarly put $x=\frac{3 y-4}{2}$ in Eq. (ii), we get

$ \begin{aligned} & & 25\left(\frac{3 y-4}{2}\right)^2+9 y^2 & =225 \\ & \Rightarrow & 225 y^2-600 y+400+36 y^2 & =900 \\ & \Rightarrow & 291 y^2-600 y-900 & =0 \\ & \Rightarrow & y_1+y_2=\frac{600}{291} & =\frac{200}{97} \end{aligned} $

Let $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ So, mid-point of $A$ and $B$

$ \begin{aligned} & \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \equiv\left(\frac{-8}{29}, \frac{100}{97}\right) \equiv(\alpha, \beta) \\ & \Rightarrow \quad \alpha=\frac{-8}{29}, \beta=\frac{100}{97} \\ & \therefore \quad 3 \beta-2 \alpha=\frac{100}{29}+\frac{16}{29}=4 \end{aligned} $

Statement I : Given, equation of ellipse is

$ \begin{aligned} & & 4 x^2+y^2-8 x-4 y+4 & =0 \\ & & 4 x^2-8 x+4+y^2-4 y+4 & =4 \\ & \Rightarrow & 4(x-1)^2+(y-2)^2 & =4 \\ & \Rightarrow & \frac{(x-1)^2}{1}+\frac{(y-2)^2}{4} & =1 \end{aligned} $

The origin of the vertex is $(1,2)$.

Here, $a^2=4, b^2=1$

$ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} $

Equation of directrix is

$ \begin{aligned} & \,\,\,\,\,\, y-k= \pm \frac{a}{e} \\ \Rightarrow &\,\,\,\,\,\, y-2= \pm \frac{2}{\sqrt{3} / 2} \\ \Rightarrow & \,\,\,\,\,\, y-2= \pm \frac{4 \sqrt{3}}{3} \\ \Rightarrow & \,\,\,\,\,\, 3 y-6= \pm 4 \sqrt{3} \Rightarrow 3 y=6 \pm 4 \sqrt{3} \end{aligned} $

∴ Statement I is true.

Statement II : Given, equation of ellipse

$ \begin{array}{rlrl} & x^2+4 y^2-4 x-8 y+4 =0 \\ \Rightarrow & x^2-4 x+4+4 y^2-8 y+4 =4 \\ \Rightarrow & (x-2)^2+4(y-1)^2 =4 \\ & \Rightarrow \frac{(x-2)^2}{4}+\frac{(y-1)^2}{1} =1 \end{array} $

$\therefore \frac{X^2}{4}+\frac{Y^2}{1}=1$, where $X=x-2$ and

$ \begin{aligned} & Y=y-1 \\ & \qquad e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} \end{aligned} $

$ \begin{aligned} \text { Coordinate of focus } & =( \pm a e, 0) \\ & =( \pm \sqrt{3}, 0) \end{aligned} $

Equation of latusrectum

$ \begin{array}{rlrl} & X = \pm \sqrt{3} \\ \Rightarrow & x-2 = \pm \sqrt{3} \\ \Rightarrow & x =2 \pm \sqrt{3} \end{array} $

∴ Statement II is false.

Given, equation of ellipse is

$ \begin{aligned} \frac{x^2}{9}+\frac{y^2}{4} & =1 \\ e & =\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3} \end{aligned} $

Focus of ellipse which lying on positive $X$-axis is (ae, 0 )

$ =\left(\frac{\sqrt{5}}{3} \times 3,0\right)=S(\sqrt{5}, 0) $

Coordinate of $P$ is $(3 \cos \theta, 2 \sin \theta)$

$ \begin{array}{cc} & S P=1 \\ \Rightarrow & (S P)^2=1 \\ & (\sqrt{5}-3 \cos \theta)^2+(2 \sin \theta-0)^2=1 \\ \Rightarrow & 5+9 \cos ^2 \theta-6 \sqrt{5} \cos \theta+4 \sin ^2 \theta=1 \\ \Rightarrow & 5+9 \cos ^2 \theta-6 \sqrt{5} \cos \theta+4 -4 \cos ^2 \theta=1 \\ \Rightarrow & 9+5 \cos ^2 \theta-6 \sqrt{5} \cos \theta=1 \\ \Rightarrow & (3-\sqrt{5} \cos \theta)^2=1 \\ \Rightarrow & 3-\sqrt{5} \cos \theta=1 \end{array} $

[ ∵ s lies on positive $X$-axis]

$ \Rightarrow \quad \sqrt{5} \cos \theta=2 $

Given, equation of ellipse is

$ \frac{x^2}{15 / a}+\frac{y^2}{15 / b}-1 $

Here, $A^2=\frac{15}{a}, B^2=\frac{15}{b}$

Distance between foci = 2

$ \begin{aligned} 2 A e & =2 \\ A e & =1 \Rightarrow e=1 / A \end{aligned} $

Distance between 2 directrix $=5$

$ \begin{aligned} \frac{2 A}{e} & =5 \Rightarrow 2 A^2=5 \\ A^2 & =\frac{5}{2} \Rightarrow \frac{15}{a}-\frac{5}{2} \\ 5 a & =30 \Rightarrow a=6 \\ e^2 & =\frac{1}{A^2}=\frac{2}{5} \Rightarrow 1-\frac{B^2}{A^2}=\frac{2}{5} \\ 1-\frac{15 / b}{15 / a} & =\frac{2}{5} \Rightarrow 1-\frac{a}{b}=\frac{2}{5} \\ \frac{a}{b} & =\frac{3}{5} \Rightarrow \frac{6}{b}=\frac{3}{5} \quad\{\because a=6\} \\ b & =10 \Rightarrow a+b=6+10=16 \end{aligned} $

Assertion vertex of the parabola is $(0,0)$.

Here $a^2=25, b^2=16$

The image of ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ with respect to line $x+y=10$ is

$ \frac{(x-h)^2}{16}+\frac{(y-k)^2}{25}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The mid-point of centres of both ellipse must lie on line $x+y=10$

$ \begin{array}{ll} \Rightarrow & \frac{h+k}{2}=10 \\ \Rightarrow & h+k=20 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Slope of the line joining the centres $=k / h$

Slope of the given line $=-1$

∴ The line joining the centres and the given line is perpendicular

$ \begin{aligned} & & \frac{k}{h}(-1) & =-1 \\ \Rightarrow & & h & =k \\ & & h+h & =20 \,\,\,\,\,\,\text { ( ∵ from Eq. (ii)) }\\ & & 2 h & =20 \\ \Rightarrow & & h & =10 \end{aligned} $

From Eq. (i), $\frac{(x-10)^2}{16}+\frac{(y-10)^2}{25}=1$

Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.

Given, $4 x^2+9 y^2=36$

$ \frac{x^2}{9}+\frac{y^2}{4}=1 $

Here, $a^2=9$ and $b^2=4$

Equation of normal is

$ \begin{aligned} & 3 x \sec \theta-2 y \operatorname{cosec} \theta=9-4 \\ & \Rightarrow \quad 3 x \sec \frac{7 \pi}{4}-2 y \operatorname{cosec} \frac{7 \pi}{4}=5 \\ & \Rightarrow \quad 3 x \times \sqrt{2}+2 y \sqrt{2}=5 \\ & \Rightarrow \quad 3 \sqrt{2} x+2 \sqrt{2} y-5=0 \end{aligned} $

$P(a \cos \theta, b \sin \theta)$

and $\quad Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$

satisfy $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$

$\therefore \frac{a^2}{\alpha^2} \cos ^2 \theta+\frac{b^2}{\beta^2} \sin ^2 \theta=1$

and $\frac{a^2}{\alpha^2} \sin ^2 \theta+\frac{b^2}{\beta^2} \cos ^2 \theta=1$

∴ Adding both equations, we get

$\frac{a^2}{\alpha^2}\left(\cos ^2 \theta+\sin ^2 \theta\right)+\frac{b^2}{\beta^2}\left(\sin ^2 \theta+\cos ^2 \theta\right)=2$

If $S \equiv(a e, 0)$, then

$ \tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}=\frac{e-1}{e+1} $

If $S \equiv(-a e, 0)$, then $\tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}=\frac{e+1}{e-1}$

$ \Rightarrow \quad\left|\frac{e-1}{e+1}\right|<1 \text { and }\left|\frac{e+1}{e-1}\right|>1 $

According to the question,

$ \begin{aligned} & \tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}=-(2 \sqrt{2}+3) \\ \Rightarrow \quad & \left|\tan \frac{\theta_1}{2} \cdot \tan \frac{\theta_2}{2}\right|=2 \sqrt{2}+3>1 \\ \Rightarrow \quad & \frac{e+1}{e-1}=-\left(2 \sqrt{2}+3 \Rightarrow e=\frac{1}{\sqrt{2}}\right. \end{aligned} $

In this case, $S \equiv(-a e, 0)$

i.e. $S \equiv\left(\frac{-a}{\sqrt{2}}, 0\right)$

Axes are rotated through $\frac{\pi}{4}$ about origin in positive direction.

$ \begin{aligned} & x \rightarrow x \cos \frac{\pi}{4}-y \sin \frac{\pi}{4}, x \rightarrow \frac{1}{\sqrt{2}}(x-y) \\ & y \rightarrow x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}, x \rightarrow \frac{1}{\sqrt{2}}(x+y) \end{aligned} $

So, equation $49 x^2+25 y^2=1225$ will transform into

$ \begin{aligned} & & \frac{49}{2}(x-y)^2+\frac{25}{2}(x+y)^2 & =1225 \\ \Rightarrow & & 37 x^2-24 x y+37 y^2 & =1225 \end{aligned} $

On comparing with $p x^2+q x y+r y^2=t$

$ p=37, q=-24, r=37, t=1225 $

$ \begin{array}{r} (p+q+r-15)^2=(37-24+37-15)^2 \\ =(35)^2=1225 \end{array} $

So, $(p+q+r-15)^2=t$

Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$ \begin{gathered} \text { Eccentricity }=\frac{\sqrt{3}}{2} \Rightarrow \sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2} \\ \begin{array}{c} 1-\frac{b^2}{a^2}=\frac{3}{4} \Rightarrow \frac{b^2}{a^2}=\frac{1}{4} \\ b^2=a^2 / 4 \end{array} \end{gathered} $

Length of latusrectum = 1

$ \frac{2 b^2}{a}=1 \quad b^2=\frac{a}{2} $

From Eqs (i) and (ii),

$ \begin{aligned} \frac{a^2}{4} & =\frac{a}{2} \Rightarrow a^2-2 a \\ a & =0 \\ a \neq 0 \quad b & =1 \end{aligned} $

Sum of length of major-axis and minor-axis

$ =2 a+2 b=4+2=6 $

Given, Focii of ellipse are $f_1(-3,0)$ and $f_2(9,0)$.

Centre of ellipse will be at mid-point of $f_1$ and $f_2$

$ \begin{aligned} & O\left(\frac{-3+9}{2}, \frac{0+0}{2}\right) \equiv O(3,0) \\ & 2 a e=12 \Rightarrow a e=6 \quad a \times \frac{1}{3}=6 \\ & \Rightarrow a=18 \text { and } \sqrt{1-\frac{b^2}{a^2}}=\frac{1}{3} \Rightarrow 1-\frac{b^2}{a^2}=\frac{1}{9} \\ & \Rightarrow \frac{b^2}{a^2}=\frac{8}{9} \Rightarrow b^2=\frac{8}{9} \times(18)^2=288 \\ & b=12 \sqrt{2} \end{aligned} $

$ \begin{aligned} &\text { Equation of ellipse : }\\ &\begin{aligned} & \frac{(x-3)^2}{(18)^2}+\frac{(y-0)^2}{288}=1 \\ & x-3=a \cos \theta, y-0=b \sin \theta \\ & x-3=18 \cos \theta, y=12 \sqrt{2} \sin \theta \\ & x=3+18 \cos \theta, y=12 \sqrt{2} \sin \theta \end{aligned} \end{aligned} $

Apply pythagoras theorem, in $\triangle A S_1 S_2$

$\begin{array}{l} & \left(S_1 A\right)^2+\left(A S_2\right)^2 =\left(S_1 S_2\right)^2 \\ & a^2+a^2 =(2 a e)^2 \\ \Rightarrow & 2 a^2 =4 a^2 e^2 \\ & \therefore e =\frac{1}{\sqrt{2}} \end{array} $

Given, equation of ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$

Thus, $a^2=4, b^2=9$

Since, $b>a$.

So, the major axis of ellipse is $Y$-axis

$\begin{aligned} e^2 & =1-\frac{a^2}{b^2}=1-\frac{4}{9}=\frac{5}{9} \\ e & =\frac{\sqrt{5}}{3} \end{aligned}$

Thus, foci of ellipse be $(0, \pm b e)=(0, \pm \sqrt{5})$

So, $S=(0, \sqrt{5})$ and $s^{\prime}=(0,-\sqrt{5})$

Given, point on ellipse is $p\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$.

Thus, focal distance

$\begin{aligned} P S & =\sqrt{\left(\frac{4}{\sqrt{5}}-0\right)^2+\left(\frac{3}{\sqrt{5}}-\sqrt{5}\right)^2}=\sqrt{\frac{4^2}{5}+\frac{4}{5}} \\ & =\sqrt{\frac{16+4}{5}}=\sqrt{4}=2 \end{aligned}$

$\begin{aligned} \text { and focal distance } P S^{\prime} & =\sqrt{\left(\frac{4}{\sqrt{5}}-0\right)^2+\left(\frac{3}{\sqrt{5}}+\sqrt{5}\right)^2} \\ & =\sqrt{\frac{16+64}{5}}=\sqrt{\frac{80}{5}} \\ & =\sqrt{16}=4 \end{aligned}$

Thus, required focal distances are 4, 2.

Let the coordinates of ends of rod be $A(a, 0), B(0, b)$.

$A B=r \Rightarrow a^2+b^2=r^2 \quad \text{... (i)}$

Let the mid-point of $A B$ be $(h, k)$

$\begin{aligned} & h=\frac{a+0}{2}=\frac{a}{2}, k=\frac{0+b}{2}=\frac{b}{2} \\ & \Rightarrow \quad a=2 h, b=2 k \\ \end{aligned}$

From Eq. (i), $(2 h)^2+(2 k)^2=r^2$

$\begin{aligned} & \Rightarrow h^2+k^2=\left(\frac{r}{2}\right)^2 \\ & \begin{aligned} \text { Length } & =\text { Circumference of circle } \\ & =2 \pi \text { (radius) } \\ & =2 \pi\left(\frac{r}{2}\right)=\pi r \end{aligned} \end{aligned}$