Ellipse

Replace $x$ with $\frac{x-y}{\sqrt{2}}$ and $y$ with $\frac{y+x}{\sqrt{2}}$ is equation

$ \begin{aligned} & \Rightarrow a x^2+2 h x y+b y^2=c \\ & \begin{aligned} \Rightarrow a\left(\frac{x-y}{\sqrt{2}}\right)^2+2 h\left(\frac{y+x}{\sqrt{2}}\right) & \left(\frac{-y+x}{\sqrt{2}}\right) +b\left(\frac{y+x}{\sqrt{2}}\right)^2=c \end{aligned} \end{aligned} $

$ \begin{aligned} a \frac{\left(x^2+y^2-2 x y\right)}{2}+ & \frac{2 h\left(-y^2+x^2\right)}{2} +\frac{b\left(y^2+x^2+2 x y\right)}{2}=c \end{aligned} $

$ \begin{aligned} \Rightarrow \frac{x^2(a+2 h+b)}{2}+ & \frac{y^2(a-2 h+b)}{2} +x y(-a+b)=c \end{aligned} $

$ \begin{aligned} & \Rightarrow a+2 h+b=25 \times 2 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, c=225 \\ & \Rightarrow a-2 h+b=9 \times 2 \\ & \Rightarrow(a+2 h+b-\sqrt{c})^2=(35)^2=1225 \end{aligned} $

$ \begin{gathered} r=\frac{a}{3}\left(\cos \theta_1+\cos \theta_2+\cos \theta_3\right) \\ s=\frac{b}{3}\left(\sin \theta_1+\sin \theta_2+\sin \theta_3\right) \\ \Rightarrow \sum_{i=1}^3 \cos \theta_i=\frac{3 r}{a} \text { and } \sum_{i=1}^3 \sin \theta_i=\frac{3 s}{b} \\ \Rightarrow\left(\sum_{i=1}^3 \cos \theta_i\right)^2+\left(\sum_{i=1}^3 \sin \theta_i\right)^2=\frac{9 r^2}{a^2}+\frac{9 s^2}{b^2} \\ \Rightarrow 3+2\left(\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)\right. \left.+\cos \left(\theta_3-\theta_1\right)\right) \\ =\frac{9 r^2}{a^2}+\frac{9 s^2}{b^2} \end{gathered} $

$ \begin{aligned} & \Rightarrow 1+\frac{2}{3}\left(\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)\right. \left.+\cos \left(\theta_3-\theta_1\right)\right)\\ & \quad=\frac{3 r^2}{a^2}+\frac{3 s^2}{b^2} \\ & \Rightarrow \frac{\left.\cos \left(\theta_3-\theta_1\right)\right)}{\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)+\cos \left(\theta_3-\theta_1\right)} \\ & =\frac{1}{2}\left(\frac{3 r^2}{a^2}+\frac{3 s^2}{b^2}-1\right) \end{aligned} $

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

Let $P \equiv\left(a t^2, 2 a t\right)$

$ \begin{aligned} & \Rightarrow \quad \frac{a^2 t^4}{a^2}+\frac{4 a^2 t^2}{2 a^2}=1 \\ & \Rightarrow \quad t^4+2 t^2=1 \end{aligned} $

Slope of tangent $P A$ to ellipse at $\left(a t^2, 2 a t\right)$

$ \begin{aligned} & \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 \\ & \frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}=0 \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)=\frac{a^2 y}{b^2 x}-\frac{b^2 x}{a^2 y} \\ &=\frac{-2 a^2 \times a t^2}{a^2 \times 2 a t}=-t \end{aligned} $

Slope of tangent $P B$ to parabola at $\left(a t^2, 2 a t\right)$

$ \begin{aligned} & y^2=4 a x \\ & 2 y \frac{d y}{d x}=4 a \\ \Rightarrow & \frac{d y}{d x}=\frac{2 a}{2 a t}=\frac{1}{t} \end{aligned} $

Product of slopes is -1 .

$ \Rightarrow \quad \theta=90^{\circ} $

Coordinates of point $\frac{2 \theta}{3}$ is

$ \begin{aligned} & \left(a \cos \left(\frac{2 \theta}{3}\right), b \sin \left(\frac{2 \theta}{3}\right)\right) \\ & \quad=\left(\frac{a}{2}, \frac{\sqrt{3}}{2} \times \sqrt{2 a}\right)=\left(\frac{a}{2}, \sqrt{\frac{3}{2}} a\right) \end{aligned} $

$ \begin{aligned} \Rightarrow & \frac{x^2}{25}+\frac{y^2}{9}=1 \\ \Rightarrow & P=(5 \cos \theta, 3 \sin \theta) \\ \Rightarrow & S S^{\prime}=2 a e=10\left(1-\frac{9}{25}\right)^{1 / 2} \\ & =10 \times\left(\frac{16}{25}\right)^{1 / 2}=10 \times \frac{4}{5} \end{aligned} $

Area of $\triangle P S^{\prime} S=\frac{1}{2} \times \frac{40}{5} \times|3 \sin \theta|$

$ \begin{aligned} & =\frac{40}{10} \times 3 \times|\sin \theta| \\ & =12|\sin \theta| \end{aligned} $

∴ Maximum area $=12$

Given equation of ellipse

$ \frac{x^2}{25}+\frac{y^2}{16}=1 $

Coordinate of one end of latus. Rectum is $\left(a e, \frac{b^2}{a}\right)$

$ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5} $

Equation of normal at $\left(5 \times \frac{3}{5}, \frac{16}{5}\right)$

Equation of normal to $\frac{x^2}{25}+\frac{y^2}{16}$ at $\left(3, \frac{16}{5}\right)$ is $y-\frac{16}{5}=m(x-3)$

$ \begin{array}{r} m=\frac{-1}{\left(\frac{d y}{d x}\right)_{3, \frac{16}{5}}} \\ \Rightarrow \quad \frac{2 x}{25}+\frac{2 y}{16} \times \frac{d y}{d x}=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{-16 x}{25 y} \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{\left(3, \frac{16}{5}\right)}=\frac{-16}{25} \times \frac{3 \times 5}{16}=\frac{-3}{5} \end{aligned} $

Equation of normal,

$ \begin{aligned} & y-\frac{16}{5}=\frac{5}{3}(x-3) \\ \Rightarrow & \frac{5 y-16}{25}=\frac{1}{3}(x-3) \\ \Rightarrow & 15 y-48=25 x-75 \\ \Rightarrow & 25 x-15 y-27=0 \end{aligned} $

Also, equation of normal given

$ \begin{aligned} & l x+m y-27=0 \\ & \frac{25}{l}=\frac{-15}{m}=+1 \\ & l=+25 \text { and } m=-15 \\ & l+m=10=\frac{6}{e}, \text { where } e=\frac{3}{5} \end{aligned} $

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

$ \begin{aligned} &\begin{aligned} \because & \frac{a}{e}-a e=\frac{4}{\sqrt{5}} \\ \Rightarrow & \frac{3}{e}-3 e=\frac{4}{\sqrt{5}} \Rightarrow 3-3 e^2=\frac{4 e}{\sqrt{5}} \\ \Rightarrow & 3 \sqrt{5} e^2+4 e-3 \sqrt{5}=0 \\ \Rightarrow & e=\frac{-4 \pm \sqrt{16+180}}{6 \sqrt{5}} \\ & =\frac{-4 \pm 14}{6 \sqrt{5}}=\frac{10}{6 \sqrt{5}}=\frac{\sqrt{5}}{3} \\ b^2 & =a^2\left(1-e^2\right) \\ & =9\left(1-\frac{5}{9}\right)=4 \end{aligned}\\ &\text { Slope of tangent at }\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\\ &\begin{aligned} \frac{d y}{d x} & =\frac{-2 x_1}{9} \cdot \frac{b^2}{2 y_1} \\ & =\frac{-2 \times \frac{3}{\sqrt{2}} \times 4}{9 \times 2 \times \frac{2}{\sqrt{2}}}=\frac{-2}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{x^2}{4}+\frac{y^2}{1}=1 \text { and } y=x+1 \\ \Rightarrow \quad & \frac{x^2}{4}+(x+1)^2=1 \\ \Rightarrow \quad & \frac{x^2}{4}+x^2+2 x=0 \\ \Rightarrow \quad & x=0 \text { and } \frac{5 x}{4}=-2 x=\frac{-8}{5} \\ & y=1, y=\frac{-3}{5} \\ \therefore & \text { Length of chord }=\sqrt{\left(\frac{8}{5}\right)^2+\left(\frac{8}{5}\right)^2}=\frac{8}{5} \sqrt{2} \end{aligned} \end{aligned} $

We have, $a=3, b=2$

$P(3 \cos \theta, 2 \sin \theta)$

Equation of normal at $P$.

$ \frac{3 x}{\cos \theta}-\frac{2 y}{\sin \theta}=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of auxiliary circle $x^2+y^2=13$

∵ Normal of circle passes through centre

$ \therefore Q=(3 \cos \theta, 3 \sin \theta) $

Normal at $Q$ to circle

$ \begin{aligned} & y=x \tan \theta \\ & \Rightarrow \quad x=y \cot \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Solving Eqs. (i) and (ii), we get

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

∴ Locus of $R$ is $x^2+y^2=25$

Let mid-point of chord be $P\left(x_1, y_1\right)$

∴ Equation of chord to ellipse $x^2+\frac{y^2}{4}=1$ is $T=S_1$

$ \begin{aligned} & x x_1+\frac{y y_1}{4}-1=x_1^2+\frac{y_1^2}{4}-1 \\ & x x_1+\frac{y y_1}{4}=x_1^2+\frac{y_1^2}{4} \end{aligned} $

Given, $x-y=-1$

$ \begin{aligned} & \frac{x_1}{1}=\frac{\frac{y_1}{4}}{-1}=\frac{x_1^2+\frac{y_1^2}{4}}{-1} \\ & y_1=-4 x_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

And $-x_1=x_1^2+\frac{y_1^2}{4}$

By Eq. (i)

$ \begin{aligned} & -x_1=x_1^2+4 x_1^2 \Rightarrow 5 x_1^2=-x_1 \\ & \therefore \quad x_1=\frac{-1}{5} \text { and } y_1=\frac{4}{5} \end{aligned} $

$ \begin{aligned} & \text { } 9 x^2+16 y^2=144 \\ & \Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1 \end{aligned} $

Let point $P$ on ellipse $=P(4 \cos \theta, 3 \sin \theta)$

Normal at $\boldsymbol{P}$

$ \begin{aligned} & \frac{4 x}{\cos \theta}-\frac{3 y}{\sin \theta}=16-9 \\ \Rightarrow \quad & \frac{4 x}{\cos \theta}-\frac{3 y}{\sin \theta}-7=0 \end{aligned} $

Distance from origin

$ \begin{aligned} D & =\left|\frac{-7}{\sqrt{\frac{16}{\cos ^2 \theta}+\frac{9}{\sin ^2 \theta}}}\right| \\ & =\left|\frac{-7}{\sqrt{16 \sec ^2 \theta+9 \operatorname{cosec}^2 \theta}}\right| \end{aligned} $

for maximum value of $D$, $P=16 \sec ^2 \theta+9 \operatorname{cosec}^2 \theta$ will be minimum

$ \frac{d P}{d \theta}=32 \sec ^2 \theta \tan \theta-18 \operatorname{cosec}^2 \theta \cot \theta=0 $

$ \begin{aligned} & \Rightarrow \quad \frac{\sin \theta}{\cos ^3 \theta}=\frac{18}{32} \cdot \frac{\cos \theta}{\sin ^3 \theta} \\ & \Rightarrow \quad \frac{\sin ^4 \theta}{\cos ^4 \theta}=\frac{9}{16} \\ & \Rightarrow \quad \tan ^2 \theta=\frac{3}{4} \Rightarrow \cot ^2 \theta=\frac{4}{3} \\ & \Rightarrow \quad \sec ^2 \theta=\frac{7}{4}, \operatorname{cosec}^2 \theta=\frac{7}{3} \\ & \quad D_{\max }=\frac{7}{\sqrt{16 \times \frac{7}{4}+9 \times \frac{7}{3}}}=\frac{7}{\sqrt{49}}=1 \end{aligned} $

$ \begin{aligned} & \text { Here, } P \equiv\left(\frac{3 a}{5}, \frac{2 b}{5}\right)=(x, y) \\ & \Rightarrow \quad x=\frac{3 a}{5}, y=\frac{2 b}{5} \\ & \Rightarrow \quad a=\frac{5 x}{3}, b=\frac{5 y}{2} \\ & \therefore \quad a^2+b^2=(15)^2 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad\left(\frac{5 x}{3}\right)^2+\left(\frac{5 y}{2}\right)^2=225 \\ & \Rightarrow \quad \frac{x^2}{9}+\frac{y^2}{4}=9 \\ & \Rightarrow \quad \frac{x^2}{81}+\frac{y^2}{36}=1 \end{aligned}\\ &\text { which represent equation of ellipse }\\ &\begin{array}{ll} \therefore & x=9 \cos \theta \\ \text { And } & y=6 \sin \theta \end{array} \end{aligned} $

Given equation of ellipse is

$ \frac{x^2}{16}+\frac{y^2}{9}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $a^2=16, b^2=9$

$ \Rightarrow \quad a=4, b=3 $

$\therefore$ Equation of tangent to the ellipse (i) is

$ \begin{aligned} & y=m x \pm \sqrt{a^2 m^2+b^2} \\ \Rightarrow \quad & y=m x \pm \sqrt{16 m^2+9} \\ \Rightarrow \quad & y-m x \mp \sqrt{16 m^2+9}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Given equation of circle $x^2+y^2=\alpha^2$

Centre $=(0,0)$, radius $=\alpha$

Since, Eq. (ii) touches the given circle

$ \begin{array}{ll} \therefore & \alpha=\left|\frac{0-0 \mp \sqrt{16 m^2+9}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & \alpha=\left|\frac{\sqrt{16 m^2+9}}{\sqrt{1+m^2}}\right| \\ \Rightarrow & \alpha^2=\frac{16 m^2+9}{1+m^2} \\ \text { Let } & f(m)=\frac{16 m^2+9}{1+m^2} \\ \Rightarrow & f(m)=\frac{16 m^2+16-16+9}{1+m^2} \\ \Rightarrow & f(m)=\frac{16\left(m^2+1\right)-7}{m^2+1} \\ & =16-\frac{7}{m^2+1} \end{array} $

Here, $m^2 \geq 0$

$ \Rightarrow \quad m^2+1 \geq 1 $

$ \begin{aligned} & \Rightarrow \quad 0 \leq \frac{1}{m^2+1} \leq 1 \\ & \Rightarrow \quad 0 \leq \frac{7}{m^2+1} \leq 7 \\ & \Rightarrow \quad 0 \geq-\frac{7}{m^2+1} \geq-7 \\ & \Rightarrow \quad 16 \geq 16-\frac{7}{m^2+1} \geq 16-7 \\ & \Rightarrow \quad 16 \geq 16-\frac{7}{m^2+1} \geq 9 \\ & \Rightarrow \quad 9 \leq f(m) \leq 16 \\ & \Rightarrow \quad 9 \leq \alpha^2 \leq 16 \\ & \Rightarrow \quad 3 \leq|\alpha| \leq 4 \end{aligned} $

The equation of ellipse is

$ \frac{x^2}{169}+\frac{y^2}{144}=1 $

So, $a^2=169, b^2=144$

$ \begin{aligned} & \Rightarrow \quad a=13, b=12 \\ & \therefore \quad c^2=a^2-b^2 \Rightarrow 169-144=25 \\ & \Rightarrow \quad c=5 \end{aligned} $

So, the foci are at $( \pm c, 0)$. So, $S=(5,0)$ and $S^{\prime}=(-5,0)$

Since, the point $B$ is one end of the minor axis lying on the positive $Y$-axis.

So, $B=(0, \pm b)=(0,12)$

$\therefore$ The vertices of the triangle are $S(5,10), S^{\prime}(-5,0)$ and $B(0,12)$

$ \begin{aligned} S S^{\prime} & =\sqrt{(-5-5)^2+(0-0)^2} \\ & =\sqrt{(-10)^2}=10 \\ S B & =\sqrt{(0-5)^2+(12-0)^2} \\ & =\sqrt{25+144}=\sqrt{169}=13 \\ S^{\prime} B & =\sqrt{(0-(-5))^2+(12-0)^2} \\ & =\sqrt{25+144}=\sqrt{169}=13 \end{aligned} $

Since, $S B=S^{\prime} B$, So $S B S^{\prime}$ is an isosceles triangle.

Now, the incenter ( $I_x, I_y$ ) of a triangle with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)$ and opposite side lengths $a, b$ and $c$ is

$ I_x=\frac{a x_1+b x_2+c x_3}{a+b+c}, I_y=\frac{a y_1+b y_2+c y_3}{a+b+c} $

Here, $\left(x_1, y_1\right)=S(5,0),\left(x_2, y_2\right)=S^{\prime}(-5,0)$, $\left(x_3, y_3\right)=B(0,12)$ and $a=S^{\prime} B=13$, $b=S B=13, C=S S^{\prime}=10$

$ \begin{array}{ll} \therefore & I_x=\frac{13(5)+13(-5)+10(0)}{13+13+10} \\ \Rightarrow & \frac{65-65+0}{36}=0 \\ & I_y=\frac{13(0)+13(0)+10(12)}{13+13+10} \\ \Rightarrow & \frac{0+0+120}{36}=\frac{10}{3} \end{array} $

So, the incenter of the $\Delta S B S^{\prime}$ is $\left(0, \frac{10}{3}\right)$.

Let the given focus be $F_1(2,-3)$ and the corresponding directrix be $L_1: 2 x+y-5=0$ and its eccentricity,

$ e=\frac{\sqrt{5}}{3} $

Let the other focus be $F_2\left(x_2, y_2\right)$

Now, the distance from $F_1(2,-3)$ to $2 x+y-5=0$ is

$ d_1=\frac{|2(2)+(-3)-5|}{\sqrt{2^2+1^2}} $

$ =\frac{|4-3-5|}{\sqrt{4+1}}=\frac{|-4|}{\sqrt{5}}=\frac{4}{5} $

Let $P(x, y)$ be any point on the ellipse.

So, $P F_1=\sqrt{(x-2)^2+(y+3)^2}$

And the distance from $P$ to the directrix $L_1$ is

$ \begin{aligned} & P M_1=\frac{|2 x+y-5|}{\sqrt{2^2+1^2}} \\ \Rightarrow \quad & \frac{|2 x+y-5|}{\sqrt{5}} \end{aligned} $

Since, $P F_1=e \cdot P M_1$

$ \begin{aligned} & \Rightarrow \sqrt{(x-2)^2+(y+3)^2} \\ & \quad=\frac{\sqrt{5}}{3} \cdot \frac{|2 x+y-5|}{\sqrt{5}} \\ & \Rightarrow \frac{|2 x+y-5|}{3} \end{aligned} $

Squaring on both sides, we get

$ \begin{aligned} &(x-2)^2+(y+3)^2=\frac{(2 x+y-5)^2}{9} \\ & \Rightarrow 9\left((x-2)^2+(y+3)^2\right) \\ &= 4 x^2+y^2+25+4 x y-20 x-10 y \\ &= 9 x^2-36 x+36+9 y^2+54 y+81 \\ &= 4 x^2+y^2+25+4 x y-20 x-10 y \\ & \Rightarrow 5 x^2+8 y^2-4 x y-16 x+64 y+92=0 \end{aligned} $

This is equation of ellipse.

Now, slope of directrix is $m_D=-2$ and the line joining the two foci is perpendicular to the directrix.

So, $m_F=\frac{-1}{m_D}=\frac{-1}{-2}=\frac{1}{2}$

Let, the other focus be $F_2\left(x_2, y_2\right)$

$ \begin{aligned} & \text { So, } \frac{y_2-(-3)}{x_2-2}=\frac{y_2+3}{x_2-2}=\frac{1}{2} \\ & \Rightarrow 2\left(y_2+3\right)=x_2-2 \\ & \Rightarrow \quad x_2=2 y_2+8 \end{aligned} $

We know that $c=a e$ and the distance between the focus and directrix is

$ \begin{aligned} & \frac{a}{e}-c=\frac{a}{e}-a e=\frac{a\left(1-e^2\right)}{e} \\ & \Rightarrow \quad \frac{a\left(1-\left(\frac{\sqrt{5}}{3}\right)^2\right)}{\frac{\sqrt{5}}{3}}=\frac{4}{\sqrt{5}} \\ & \Rightarrow \quad \frac{a\left(1-\frac{5}{9}\right)}{\frac{\sqrt{5}}{3}}=\frac{4}{\sqrt{5}} \Rightarrow a\left(\frac{4}{9}\right) \times \frac{3}{\sqrt{5}}=\frac{4}{\sqrt{5}} \\ & \Rightarrow \quad \frac{4 a}{3 \sqrt{5}}=\frac{4}{\sqrt{5}} \Rightarrow a=3 \\ & \therefore \quad c=a e=3 \times \frac{\sqrt{5}}{3}=\sqrt{5} \end{aligned} $

The distance between two foci $=2 c=2 \sqrt{5}$

Let $F_2\left(x_2, y_2\right)$. So, the distance

$ F_1 F_2=\sqrt{\left(x_2-2\right)^2+\left(y_2+3\right)^2}=2 \sqrt{5} $

$ \begin{aligned} & \Rightarrow \sqrt{\left(2 y_2+8-2\right)^2+\left(y_2+3\right)^2}=2 \sqrt{5} \\ & \quad\left(\because x_2=2 y_2+8\right) \\ & \Rightarrow \quad \sqrt{4\left(y_2+3\right)^2+\left(y_2+3\right)^2}=2 \sqrt{5} \\ & \Rightarrow \quad \sqrt{5\left(y_2+3\right)^2}=2 \sqrt{5} \\ & \Rightarrow \quad \sqrt{5}\left|y_2+3\right|=2 \sqrt{5} \Rightarrow\left|y_2+3\right|=2 \end{aligned} $

So, $y_2+3=2$ or $y_2+3=-2$

$ \begin{array}{ll} \Rightarrow \quad & y_2=-1 \text { or } y_2=-5 \\ \therefore & x_2=2(-1)+8=6 \text { and } \\ & x_2=2(-5)+8=-2 \end{array} $

So, $F_2(6,-1)$ and $F_2(-2,-5)$

Line $2 x+y-5=0$

For $F_1(2,-3), 2(2)+(-3)-5=4-8=-4$

For $F_2(6,-1), 2(6)+(-1)-5=12-6=6$

Since, -4 and 6 have opposite signs, so foci of an ellipse is $F_2(-2,-5)$.

To solve the problem of finding the length of the major axis of the given ellipse, let's break it down step by step:

Given Information:

Focus of the ellipse: $ S = (-1, -1) $

Equation of the directrix: $ x + y + 1 = 0 $

Eccentricity of the ellipse: $ e = \frac{1}{\sqrt{2}} $

Equation of the Ellipse:

The equation connecting the focus, directrix, and eccentricity is:

$ PS = e \cdot PM $

where $ PS $ is the distance from a point $ P(x, y) $ on the ellipse to the focus $ S $, and $ PM $ is the perpendicular distance from $ P $ to the directrix.

Distance $ PS $:

$ PS = \sqrt{(x + 1)^2 + (y + 1)^2} $

Perpendicular distance $ PM $ from $ P $ to the directrix:

$ PM = \frac{|x + y + 1|}{\sqrt{2}} $

Substitute into ellipse equation:

$ \sqrt{(x + 1)^2 + (y + 1)^2} = \frac{1}{\sqrt{2}} \cdot \frac{|x + y + 1|}{\sqrt{2}} $

This simplifies to:

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

Form an equation by solving and simplifying:

Upon squaring both sides and simplifying, the equation becomes:

$ 3x^2 + 3y^2 - 2xy + 6x + 6y + 7 = 0 $

Finding the Center:

To determine the center of the ellipse, we take partial derivatives and solve:

Derivative with respect to $ x $:

$ \frac{\partial f}{\partial x} = 6x - 2y + 6 = 0 $

Derivative with respect to $ y $:

$ \frac{\partial f}{\partial y} = 6y - 2x + 6 = 0 $

Solving these equations simultaneously gives the center $ O \left( -\frac{3}{2}, -\frac{3}{2} \right) $.

Determine the Length of Major Axis:

Distance $ OS $ (from center to the focus):

$ OS = ae = \frac{1}{\sqrt{2}} $

Given:

$ a \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} $

This implies:

$ a = 1 $

Length of the major axis:

$ 2a = 2 \times 1 = 2 $

Hence, the length of the major axis of the ellipse is 2.

To find where the normal drawn at the point $(2, -1)$ on the ellipse $x^2 + 4y^2 = 8$ meets the ellipse again, we start by determining the equation of the normal. The given ellipse is $\frac{x^2}{8} + \frac{y^2}{2} = 1$.

The equation of the normal at the point $(2, -1)$ is derived as follows:

Calculate the slope of the normal at $(2, -1)$:

$ \frac{8x}{2} - \frac{2y}{-1} = 8 - 2 $

Simplify the equation to find the normal line:

$ 4x + 2y = 6 \quad \Rightarrow \quad 2x + y = 3 \quad \Rightarrow \quad y = 3 - 2x $

To find where this normal line meets the ellipse again, substitute $y = 3 - 2x$ into the ellipse equation:

$ x^2 + 4(3 - 2x)^2 = 8 $

Expand and simplify the equation:

$ x^2 + 4(9 - 12x + 4x^2) = 8 $

$ x^2 + 36 - 48x + 16x^2 = 8 $

Combine and equate:

$ 17x^2 - 48x + 28 = 0 $

Factorize:

$ (17x - 14)(x - 2) = 0 $

The solutions for $x$ are $x = 2$ and $x = \frac{14}{17}$.

Choosing $a = \frac{14}{17}$, we find:

$ 17a = 14 $

To find the area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes, we can follow these steps:

Identify Intercepts:

The x-intercept is found by setting $y = 0$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$. This gives $x = a$.

The y-intercept is found by setting $x = 0$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$. This results in $y = b$.

Vertices of the Triangle:

The vertices of the triangle formed by this line with the axes are $(a, 0)$, $(0, b)$, and the origin $(0, 0)$.

Calculate the Area:

The area of a triangle formed by vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

$ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| $

Substituting the vertices $(a, 0)$, $(0, b)$, and $(0, 0)$, we get:

$ \text{Area} = \frac{1}{2} \left| a(0-b) + 0(0-0) + 0(b-0) \right| = \frac{1}{2} | -ab | = \frac{ab}{2} $

Substitute Known Values:

Since we've identified from the given equations that $a = 1$ and $b = 7$, substitute these values into the area formula:

$ \text{Area} = \frac{1 \cdot 7}{2} = \frac{7}{2} $

Therefore, the area of the triangle formed by the line with the coordinate axes is $\frac{7}{2}$.

Slope of line $S S^{\prime}=\frac{-3}{2}$

Equation of $S S^{\prime} \Rightarrow(y-1)=\frac{-3}{2}(x+1)$

$ \Rightarrow \quad 3 x+2 y+1=0 $

Now, point of intersection of

$ \begin{aligned} & 3 x+2 y+1=0 \text { and } 2 x-3 y+1=0 \text { is } \\\\ & S^{\prime} \equiv\left(\frac{-5}{13}, \frac{1}{13}\right) \end{aligned} $

Let length of major axis $=P$

Now, $O S^{\prime}=\frac{P}{e}$ and $O S=P e$

$ \begin{aligned} \Rightarrow & O S^{\prime}-O S=\sqrt{\left(-1+\frac{5}{13}\right)^2+\left(1-\frac{1}{13}\right)^2} \\\\ \Rightarrow & \frac{P}{e}-P e=\sqrt{\left(\frac{-13+5}{13}\right)^2+\left(\frac{13-1}{13}\right)^2} \\\\ & =\sqrt{\frac{64}{169}+\frac{144}{169}}=\sqrt{\frac{208}{169}} \end{aligned} $

Given:

The equation for the latus rectum of an ellipse is $\frac{2b^2}{a} = 4$.

The distance between the foci is given as $4\sqrt{2}$.

Let's solve step-by-step:

From the latus rectum condition, we have $b^2 = 2a$.

The distance between the foci of the ellipse is defined as $2ae = 4\sqrt{2}$.

$ 2ae = 4\sqrt{2} $

Squaring both sides, we get:

$ 4a^2e^2 = 32 $

We know the relationship between the eccentricity and the axes is $e^2 = 1 - \frac{b^2}{a^2}$.

$ a^2\left(1 - \frac{b^2}{a^2}\right) = 8 $

Substituting $b^2 = 2a$ in the above equation:

$ a^2 - 2a = 8 $

Solving for $a$, we get:

$ a^2 - 2a - 8 = 0 $

Solving this quadratic equation gives $a = 4$.

Subsequently, $b^2 = 2a = 8$.

Finally, calculating $a^2 + b^2$:

$a^2 = 16$

$b^2 = 8$

Therefore, $a^2 + b^2 = 16 + 8 = 24$.

To determine the curve on which the points $(a, b)$ lie, we begin by considering the properties of the ellipse given by $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where $a > b$.

The extremities of the latus rectum of an ellipse for the positive ordinate are $\left(ae, \frac{b^{2}}{a}\right)$, where $e$ is the eccentricity of the ellipse. For an ellipse, the relationship between the semi-major axis $a$, semi-minor axis $b$, and eccentricity $e$ is given by:

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

Since these extremities lie on the parabola described by the equation $x^{2} + 2ay = 4$, we substitute the point $\left(ae, \frac{b^{2}}{a}\right)$ into this equation:

$ (ae)^{2} + 2a\left(\frac{b^{2}}{a}\right) = 4 $

This simplifies to:

$ a^{2}e^{2} + 2b^{2} = 4 $

Using the relationship for eccentricity, we substitute $a^{2}e^{2} = a^{2} - b^{2}$ into the equation:

$ a^{2} - b^{2} + 2b^{2} = 4 $

This further simplifies to:

$ a^{2} + b^{2} = 4 $

Thus, the points $(a, b)$ lie on the curve:

$ x^{2} + y^{2} = 4 $

Replacing $(a, b)$ with $(x, y)$, we confirm that the correct description for this relationship is given by:

$ x^{2} + y^{2} = 4 $

Length of latus rectum $ =\frac{2 b^{2}}{a}=\frac{8}{3} $

$ \Rightarrow \quad b^{2}=\frac{4 a}{3} $

Also, given, $ a e=\sqrt{5} $

On squaring both sides, we get

$ \begin{array}{c} a^{2} e^{2}=5 \\\\ a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=5 \Rightarrow a^{2}-\frac{4 a}{3}=15 \\\\ 3 a^{2}-4 a=15 \\\\ 3 a^{2}-4 a-15=0 \\\\ a=3 \quad a=\frac{-5}{3} \end{array} $

If $ a=3 $,

then, $ b=2 $

then, $ \sqrt{9+4+36}=\sqrt{49}=7 $

$ \frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1, s^{\prime}=(-5 e, 0) $

$ 2 \times 5 \times e=8 $

$ e=\frac{4}{5} \Rightarrow e^{2}=1-\frac{b^{2}}{a^{2}} \Rightarrow \frac{16}{25}=\frac{25-b^{2}}{25} $

$ \Rightarrow \quad b=3 $

$ \because \quad S^{\prime} \equiv(-4,0) $

$ P=(5 \cos \theta, 3 \sin \theta) $

$ \mathrm{S}^{\prime} \mathrm{P}=7 $

$ =\sqrt{(5 \cos \theta+4)^{2}+9 \sin ^{2} \theta}=49 $

On squaring both side, we get

$ \begin{array}{cc} \Rightarrow & 49=25 \cos ^{2} \theta+40 \cos \theta \\\\ & +16+9 \sin ^{2} \theta \\\\ \Rightarrow & 49=25 \cos ^{2} \theta+40 \cos \theta \\\\ & +16+9\left(1-\cos ^{2} \theta\right) \\\\ \Rightarrow & 16 \cos ^{2} \theta+40 \cos \theta-24=0 \\\\ \Rightarrow & \cos \theta=\frac{1}{2} \text { or } \cos \theta=-3 \\\\ \therefore & \theta=\frac{\pi}{3} \end{array} $

Given, ellipse :

$ 9 x^2+4 y^2-18 x-16 y-11=0 $

The given equation of ellipse can be written as

$ \frac{(x-1)^2}{2^2}+\frac{(y-2)^2}{3^2}=1 $

Here, $a=2$ and $b=3, h=1$ and $k=2$

So, the given ellipse is a vertical ellipse.

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

$\therefore$ Equations of directrices will be

$ y=k \pm \frac{b}{e} \text { i.e. } y=2 \pm \frac{9}{\sqrt{5}} $

$ \text { Ellipse: } \frac{(x-1)^2}{2^2}+\frac{(y-2)^2}{3^2}=1 $

Given, ellipse $3 x^2+4 y^2=12$

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

Latus rectums $x= \pm a e= \pm(2)\left(\frac{1}{2}\right)= \pm 1$

$\because L_1^{\prime}$ is the end of a latus rectum and it is lying in the third quadrant

$ \because \quad L_1^{\prime}=\left(-1,-\sqrt{3\left(1-\frac{(-1)^2}{4}\right)}\right)=\left(-1,-\frac{3}{2}\right) $

Equation tangent at $\left(-1, \frac{-3}{2}\right)$ is given by

$ \frac{x(-1)}{4}+\frac{y\left(-\frac{3}{2}\right)}{3}=1 \Rightarrow x+2 y+4=0 $

Slope of tangent $=-\frac{1}{2}$

$\therefore$ Slope of normal $=-\left(\frac{1}{-\frac{1}{2}}\right)=2$. Equation of normal from $\left(-1,-\frac{3}{2}\right)$ is given by

$ y+\frac{3}{2}=2(x+1) \Rightarrow 4 x-2 y+1=0 $

The points of intersection of the normal

$ 4 x-2 y+1=0 $

and the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ are

$ \begin{aligned} &L_1^{\prime}\left(-1,-\frac{3}{2}\right) \text { and } P\left(\frac{11}{19}, \frac{63}{38}\right)\\\\ &\text { Hence, } a=\frac{11}{19} \end{aligned} $

We have, $x^2+3 y^2=K$

$ \Rightarrow \frac{x^2}{(\sqrt{K})^2}+\frac{y^2}{\left(\sqrt{\frac{K}{3}}\right)^2}=1 $

So, equation of normal is

$ \sqrt{K} x \sec \theta-\sqrt{\frac{K}{3}} y \operatorname{cosec} \theta=K-\frac{K}{3} \quad \ldots \text { (i) } $

Given equation of normal is

$ 6 x-5 y-20=0 \quad \ldots \text { (ii) } $

$\because$ Eqs, (i) and (ii) represent the same line

$ \begin{aligned} & \therefore \frac{\sqrt{K} \sec \theta}{6}=\frac{\sqrt{\frac{K}{3}} \operatorname{cosec} \theta}{5}=\frac{K-\frac{K}{3}}{20} \\\\ & \Rightarrow \frac{\sqrt{K} \sec \theta}{6}=\frac{\sqrt{K} \operatorname{cosec} \theta}{5 \sqrt{3}}=\frac{K}{30} \\\\ & \text { So, } \cos \theta=\frac{5}{\sqrt{K}}, \sin \theta=\frac{2 \sqrt{3}}{\sqrt{K}} \\\\ & \because \cos ^2 \theta+\sin ^2 \theta=1 \\\\ & \quad \frac{25}{K}+\frac{12}{K}=1 \\\\ & \Rightarrow \quad K=37 \end{aligned} $

$ \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} $

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} $