Ellipse

Given

• Center of ellipse : $(1,-2)$

• Focus : $(3,-2)$

• Vertex : $(5,-2)$

The center, focus, and vertex all have the same $y$-coordinate $(-2)$, which means the major axis of the ellipse is horizontal (parallel to the $x$-axis).

Find $a$ (semi-major axis)

Distance from center to vertex :

$ a=|5-1|=4 $

Find $c$ (distance from center to focus)

$ c=|3-1|=2 $

Find $b$ using ellipse relation

$ c^2=a^2-b^2 $

$\Rightarrow $ $ 2^2=4^2-b^2 $

$\Rightarrow 4=16-b^2 $

$\Rightarrow b^2=12 $

$\Rightarrow b=2 \sqrt{3}$

Length of latus rectum

For an ellipse :

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

$=\frac{2(12)}{4}=6 $

length of latus rectum of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ is given as:

$ l=\frac{2 b^2}{a}=30 \Rightarrow b^2=15 a $

It is given that eccentricity is the maximum value of the function $f(t)=-t^2+2 t-\frac{3}{4}$

max value of quadratic function $a x^2+b x+c$ is given as $\frac{-D}{4 a}=\frac{4 a c-b^2}{4 a}$

$ f(t)_{\max }=\frac{4(-1)\left(-\frac{3}{4}\right)-(2)^2}{4(-1)}=\frac{3-4}{-4}=\frac{1}{4} $

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

squaring

$ 1-\frac{b^2}{a^2}=\frac{1}{16} $

put value of $b^2=15 a$

$ \begin{aligned} & \frac{a^2-15 a}{a^2}=\frac{1}{16} \\ & 16 a^2-240 a=a^2 \\ & 15 a^2-240 a=0 \\ & a(15 a-240)=0 \\ & \text { since } a \neq 0, a=\frac{240}{15}=16 \end{aligned} $

Now, value of $a^2+b^2=a^2+15 a\left\{b^2=15 a\right\}$

$ \begin{aligned} & =(16)^2+15 \times 16 \\ & =256+240=496 \end{aligned} $

Given, for ellipse $E_1$ and $E_2$ eccentricity is $\frac{4}{5}$.

$ E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad a>b $

distance between foci $=8$

$ 2 a e=8 \Longrightarrow a e=4 $

$\Rightarrow $ $a \times \frac{4}{5}=4 \Longrightarrow a=5$

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

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

$\Rightarrow $ $\frac{b^2}{5^2}=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow $ $b=3$

Length of latus rectum for $E_1, l_1=\frac{2 b^2}{a} $

$l_1=2 \times \frac{3^2}{5}=\frac{18}{5}$

For $E_2: \frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \quad(A < B)$

$\begin{aligned} & \text { eccentricity } e=\frac{4}{5} \\ & \text { as } A < B, e=\sqrt{1-\frac{A^2}{B^2}}=\frac{4}{5}\end{aligned}$

$1-\frac{A^2}{B^2}=\frac{16}{25}$

$\Rightarrow $ $\frac{A^2}{B^2}=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow $ $ A^2=\frac{9}{25} B^2 $

If $l_2$ is latus rectum of $E_2, l_2=\frac{2 A^2}{B}$

$l_2=\frac{2}{B}\left(\frac{9}{25} B^2\right)=\frac{18}{25} B$

It is given that $2 l_1^2=9 l_2$

$ 2\left(\frac{18}{5}\right)^2=9 \times \frac{18}{25} B $

$\Rightarrow $ $2 \times \frac{18}{9}=B$

$\Rightarrow $ $ B=4 $

distance between foci of $E_2$ is $2 B e$

$ =2 \times 4 \times \frac{4}{5}=\frac{32}{5} $

Let the given ellipses be :

$ \begin{aligned} & S_1 \equiv x^2+2 y^2-6 x-12 y+23=0 \\ & S_2 \equiv 4 x^2+2 y^2-20 x-12 y+35=0 \end{aligned} $

The equation of a curve passing through the intersection of these two ellipses is given by the family of curves $S_1+\lambda S_2=0$ :

$ \left(x^2+2 y^2-6 x-12 y+23\right)+\lambda\left(4 x^2+2 y^2-20 x-12 y+35\right)=0 $

$\Rightarrow $$ (1+4 \lambda) x^2+(2+2 \lambda) y^2-(6+20 \lambda) x-(12+12 \lambda) y+(23+35 \lambda)=0 $

Condition for circle: coefficient of $x^2=$ coefficient of $y^2$

$ 1+4 \lambda=2+2 \lambda $

$\Rightarrow $ $2 \lambda=1$

$\Rightarrow $ $ \lambda=\frac{1}{2} $

Substitute $\lambda=\frac{1}{2}$ :

$ 3 x^2+3 y^2-16 x-18 y+\frac{81}{2}=0 $

$\Rightarrow $ $x^2+y^2-\frac{16}{3} x-6 y+\frac{27}{2}=0$

$\begin{aligned} & 2 g=-\frac{16}{3} \Longrightarrow g=-\frac{8}{3} \\ & 2 f=-6 \Longrightarrow f=-3 \\ & \text { Centre }(a, b)=(-g,-f)=\left(\frac{8}{3}, 3\right) \\ & a=\frac{8}{3}, b=3\end{aligned}$

$\begin{aligned} & r^2=g^2+f^2-c \\ & r^2=\left(-\frac{8}{3}\right)^2+(-3)^2-\frac{27}{2} \\ & r^2=\frac{64}{9}+9-\frac{27}{2} \\ & r^2=\frac{128+162-243}{18} \\ & r^2=\frac{47}{18}\end{aligned}$

$\begin{aligned} & a b+18 r^2=\left(\frac{8}{3}\right)(3)+18\left(\frac{47}{18}\right) \\ & a b+18 r^2=8+47=55\end{aligned}$

$ \begin{aligned} &\begin{aligned} & y=x+1 \text { intersects } \frac{x^2}{2}+y^2=1 \\ & \Rightarrow \frac{x^2}{2}+(x+1)^2=1 \Rightarrow x^2+2 x^2+4 x=0 \\ & \Rightarrow x=0, \frac{-4}{3} \end{aligned}\\ &\text { ⇒ Points } A \text { and } B \text { are }\\ &\begin{aligned} & (0,1),\left(\frac{-4}{3}, \frac{-1}{3}\right) \\ & \theta=\tan ^{-1}\left(\frac{\frac{-1}{3}}{\frac{-4}{3}}\right)=\tan ^{-1}\left(\frac{1}{4}\right) \\ & \Rightarrow \angle A O B=\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{4}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \because \mathrm{P} \text { lies on ellipse } \Rightarrow \frac{\alpha^2}{25}+\frac{\beta^2}{9}=1 \\ & \because P S+P S^{\prime}=2 a \Rightarrow P S+P S^{\prime}=10 \\ & \because(P S)^2+\left(P S^{\prime}\right)^2-P S . P S^{\prime}=37 \\ & \left(P S+P S^{\prime}\right)-3 P S . P S^{\prime}=37 \\ & 100-3 P S . P S^{\prime}=37 \\ & 3 P S . P S^{\prime}=63 \Rightarrow P S . P S^{\prime}=21 \\ & \because P S \& P S^{\prime} \text { are }\left(5 \pm \frac{4}{5} . \alpha\right) \\ & \therefore P S . P S^{\prime}=25-\frac{16}{25} \alpha^2=21 \\ & \frac{16}{25} \alpha^2=4 \\ & \alpha=\frac{5}{2} \Rightarrow \alpha^2=\frac{25}{4} \\ & \therefore \beta^2=\frac{27}{4} \\ & \therefore \alpha^2+\beta^2=\frac{52}{4}=13 \end{aligned} $

$ \begin{aligned} & \alpha x+4 y-\sqrt{7}=0 \text { touches } 3 x^2+4 y^2=1 \\ & \therefore c^2=a^2 m^2+b^2 \\ & \frac{7}{16}=\frac{1}{3} \times \frac{\alpha^2}{16}+\frac{1}{4} \Rightarrow \alpha=3,-3 \end{aligned} $

Tangent is $3 x+47-\sqrt{7}=0$

Let the point of contact is $P\left(x_1, y_1\right)$

$ \begin{aligned} & \therefore \frac{3 x_1}{3}=\frac{4 y_1}{4}=\frac{1}{\sqrt{7}} \\ & \therefore P\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right) \\ & \text { ecc. }=\sqrt{1-\frac{3}{4}}=\frac{1}{2} \end{aligned} $

$ \begin{aligned} & P S=e(P M)=e\left(\frac{a}{e}-\frac{1}{\sqrt{7}}\right)=\frac{1}{2}\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}}\right)=\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{7}} \\ & P S^{\prime}=e\left(P M^{\prime}\right)=\frac{1}{2}\left(\frac{a}{e}+\frac{1}{\sqrt{7}}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}} \end{aligned} $

Since the ellipse has its major axis along the $y$-axis, the denominator under $y^2$ must be bigger than the denominator under $x^2$.

So we need

$f(3a+15) > f\left(a^2+7a+3\right).$

Given that $f$ is a strictly decreasing positive function, a bigger function value happens when the input is smaller. So,

$3a+15 < a^2+7a+3.$

Bring all terms to one side:

$a^2+7a+3-3a-15>0$

$a^2+4a-12>0.$

Complete the square:

$a^2+4a-12=(a+2)^2-16.$

So the inequality becomes

$(a+2)^2-16>0 \;\Rightarrow\; (a+2)^2>16.$

Taking square roots,

$|a+2|>4 \;\Rightarrow\; a+2>4 \text{ or } a+2<-4.$

So,

$a>2 \text{ or } a<-6.$

That means the allowed values are

$a \in \mathbf{R}-[-6,2].$

Hence, $\alpha=-6$ and $\beta=2$. Therefore,

$\alpha^2+\beta^2=(-6)^2+(2)^2=36+4=40.$

Since the centre of the ellipse is at the origin and one directrix is $x=9$, the major axis is along the $x$-axis.

Let the ellipse be

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

with eccentricity

$ e=\frac{1}{3} $

For an ellipse with major axis along the $x$-axis, the directrices are

$ x=\pm \frac{a}{e} $

Given one directrix is $x=9$, so

$ \frac{a}{e}=9 $

Since $e=\frac13$,

$ a=9\cdot \frac13=3 $

Also,

$ c=ae=3\cdot \frac13=1 $

So the focus $P(\alpha,0)$ is

$ P=(1,0) $

Hence $\alpha=1$.

Now,

$ b^2=a^2(1-e^2)=9\left(1-\frac19\right)=9\cdot \frac89=8 $

Therefore the ellipse is

$ \frac{x^2}{9}+\frac{y^2}{8}=1 $

Let the midpoint of chord $AB$ passing through the focus $(1,0)$ be $(h,k)$.

For the ellipse

$ \frac{x^2}{9}+\frac{y^2}{8}=1 $

the equation of the chord whose midpoint is $(h,k)$ is given by the NCERT midpoint formula:

$ \frac{xx_1}{9}+\frac{yy_1}{8}=\frac{x_1^2}{9}+\frac{y_1^2}{8} $

So here the chord with midpoint $(h,k)$ is

$ \frac{xh}{9}+\frac{yk}{8}=\frac{h^2}{9}+\frac{k^2}{8} $

Since this chord passes through the focus $(1,0)$, substitute $(x,y)=(1,0)$:

$ \frac{h}{9}=\frac{h^2}{9}+\frac{k^2}{8} $

Multiply by $72$:

$ 8h=8h^2+9k^2 $

So,

$ 9k^2=8h-8h^2=8h(1-h) $

Replacing $(h,k)$ by $(x,y)$ for the locus of the midpoint, we get

$ 9y^2=8x(1-x) $

Hence the required locus is

$ \boxed{9y^2=8x(1-x)} $

So the correct option is A.

For the ellipse

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

a focus is $S(4,0)$ and eccentricity is

$ e=\frac{4}{5} $

We need the area of $\triangle POS$, where $P(3,\alpha)$ lies on the ellipse and $O=(0,0)$.

Since the focus is on the $x$-axis, the ellipse has major axis along the $x$-axis.

For such an ellipse, focus is at $(\pm ae,0)$.

Given one focus is $(4,0)$, so

$ ae=4 $

and $e=\frac{4}{5}$. Therefore,

$ a\cdot \frac{4}{5}=4 $

$ a=5 $

Now,

$ c=ae=4 $

and for ellipse,

$ b^2=a^2-c^2 $

So,

$ b^2=25-16=9 $

Hence the ellipse is

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

Since $P(3,\alpha)$ lies on the ellipse, substitute $x=3$:

$ \frac{3^2}{25}+\frac{\alpha^2}{9}=1 $

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

$ \frac{\alpha^2}{9}=1-\frac{9}{25} =\frac{16}{25} $

$ \alpha^2=\frac{144}{25} $

$ \alpha=\pm \frac{12}{5} $

Now find the area of triangle with vertices

$ O(0,0),\quad P(3,\alpha),\quad S(4,0) $

Using base $OS$:

$ OS=4 $

Since $OS$ lies on the $x$-axis, the height from $P$ to $OS$ is $|\alpha|$.

So area is

$ \frac{1}{2}\times 4\times |\alpha| =2|\alpha| $

$ =2\times \frac{12}{5} =\frac{24}{5} $

Therefore, the area of $\triangle POS$ is

$ \boxed{\frac{24}{5}} $

So, the correct option is:

$ \boxed{\text{Option C}} $

Centroid :

Let $Q$ be a point on the given circle. First write the circle in standard form:

$x^2+y^2-14x-14y+82=0 \Rightarrow (x-7)^2+(y-7)^2=16$

So we can take

$Q(7+4\cos\theta,\;7+4\sin\theta)$

Point $R$ lies on the line $x+y=5$. Let its ordinate be $y$, then its abscissa is $5-y$, so

$R(5-y,\;y)$

Now use the centroid formula. For triangle $PQR$, centroid is

$ \left(\frac{x_P+x_Q+x_R}{3},\;\frac{y_P+y_Q+y_R}{3}\right) $

Given $P(3\cos\alpha,\;2\sin\alpha)$, so the centroid becomes

$ \begin{aligned} & \left(\frac{3 \cos \alpha+7+4 \cos \theta+5-y}{3}, \frac{2 \sin \alpha+7+4 \sin \theta+y}{3}\right) \\ & =\left(\cos \alpha+2, \frac{2}{3} \sin \alpha+3\right) \end{aligned} $

On comparison

Compare the $x$-coordinates:

$\frac{3 \cos \alpha+7+4 \cos \theta+5-y}{3}=\cos\alpha+2 \Rightarrow 4\cos\theta=y-6$

So,

$ \cos \theta=\frac{y-6}{4} $

Compare the $y$-coordinates:

$\frac{2 \sin \alpha+7+4 \sin \theta+y}{3}=\frac{2}{3}\sin\alpha+3 \Rightarrow 4\sin\theta=2-y$

So,

$ \sin \theta=\frac{2-y}{4} $

Now use the identity $\sin^2\theta+\cos^2\theta=1$:

$\sin ^2 \theta+\cos ^2 \theta=1 \Rightarrow(y-6)^2+(y-2)^2=16$

Simplify:

$y^2-8 y+12=0$

This quadratic gives two possible values of $y$ (ordinates of $R$). The sum of roots is $8$, so

$\Rightarrow \text { Sum of ordinates }=8$

Since $a < b$, the major axis is along the $y$-axis. So for the ellipse

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

we use the eccentricity formula

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

Given

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

so

$ e^2=\frac{5}{9}. $

Hence,

$ 1-\frac{a^2}{b^2}=\frac{5}{9} $

which gives

$ \frac{a^2}{b^2}=1-\frac{5}{9}=\frac{4}{9}. $

Therefore,

$ a^2=\frac{4}{9}b^2. $

Now the ellipse passes through $(4,3)$, so it satisfies the equation:

$ \frac{4^2}{a^2}+\frac{3^2}{b^2}=1. $

That is,

$ \frac{16}{a^2}+\frac{9}{b^2}=1. $

Substitute $a^2=\frac{4}{9}b^2$:

$ \frac{16}{\frac{4}{9}b^2}+\frac{9}{b^2}=1. $

Simplifying,

$ 16\cdot \frac{9}{4b^2}+\frac{9}{b^2}=1 $

$ \frac{36}{b^2}+\frac{9}{b^2}=1 $

$ \frac{45}{b^2}=1 $

so

$ b^2=45. $

Then

$ a^2=\frac{4}{9}\cdot 45=20. $

For an ellipse with major axis along the $y$-axis, the length of latus rectum is

$ \frac{2a^2}{b}. $

So,

$ \text{Length of latus rectum}=\frac{2\cdot 20}{\sqrt{45}} =\frac{40}{3\sqrt{5}}. $

Rationalising,

$ \frac{40}{3\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}} =\frac{8\sqrt{5}}{3}. $

Therefore, the correct answer is

$ \boxed{\frac{8\sqrt{5}}{3}} $

So, Option D is correct.

$\begin{aligned} &E: \frac{x^2}{4 / 3}+\frac{y^2}{4 / P}=1\\ &\text { Centre of circle }(1,2) \text {, radius }\\ &\mathrm{r}=\sqrt{1+4+11} \end{aligned}$

$\mathrm{r}=4$

$\because$ E pass from centre $(1,2)$

$\therefore \frac{3}{4}+\mathrm{P}=1$

$\mathrm{P}=\frac{1}{4} \quad \therefore$ vertical ellipse

$\begin{aligned} & \mathrm{e}=\sqrt{1-\frac{4 / 3}{16}}=\sqrt{1-\frac{1}{12}}=\sqrt{\frac{11}{12}} \\ & \therefore \text { Focal distance of } \mathrm{C}(\mathrm{~h}, \mathrm{k}) \\ & =\mathrm{b} \pm \mathrm{ek} \\ & \mathrm{~F}_1=4+\sqrt{\frac{11}{12}} \times 2 \\ & \mathrm{~F}_2=4-\sqrt{\frac{11}{12}} \times 2 \\ & \therefore \mathrm{~F}_1 \mathrm{~F}_2=16-\frac{11}{3}=\frac{37}{3} \end{aligned}$

$\therefore 6 \mathrm{~F}_1 \mathrm{~F}_2-\mathrm{r}=74-4=70$

Given that the length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is 10, we have:

$ \frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 \tag{1} $

Next, consider the function $ f(t) = t^2 + t + \frac{11}{12} $. To find its minimum value, we calculate the derivative:

$ \frac{df(t)}{dt} = 2t + 1 = 0 \quad \Rightarrow \quad t = \frac{-1}{2} $

Plugging $t = -\frac{1}{2}$ into $f(t)$ gives the minimum value:

$ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3} $

Thus, the eccentricity $e$ of the ellipse is $\frac{2}{3}$, so $e^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

Using the eccentricity formula for an ellipse:

$ e^2 = \frac{1 - b^2}{a^2} \quad \Rightarrow \quad \frac{4}{9} = \frac{1 - b^2}{a^2} $

Rearranging gives:

$ b^2 = a^2 \left(1 - \frac{4}{9}\right) = a^2 \cdot \frac{5}{9} $

From equation $(1)$, $b^2 = 5a$. Substituting, we have:

$ 5a = a^2 \cdot \frac{5}{9} \quad \Rightarrow \quad a^2 = 9a $

Solving for $a$,

$ a = 9, \quad \text{and therefore } b^2 = 5a = 45 \quad \Rightarrow \quad b = \sqrt{45} = 3\sqrt{5} $

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

$ a^2 + b^2 = 81 + 45 = 126 $

$\begin{aligned} & \text { Total trangles }=\Rightarrow={ }^{\mathrm{h}} \mathrm{C}_3 \\ & \text { Total auadrilaterals }={ }^{\mathrm{h}} \mathrm{C}_4=\mathrm{q} \\ & { }^{\mathrm{n}} \mathrm{C}_3+{ }^{\mathrm{n}} \mathrm{C}_4=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_4=126 \\ & \Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8 \\ & \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{n}}=1 \Rightarrow \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{8}=1 \\ & \mathrm{e}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}} \end{aligned}$

$E: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$

$\begin{aligned} & T: y=m x \pm \sqrt{16 m^2+9} \\ & y=x+p \\ & \Rightarrow m=1 \\ & \Rightarrow p= \pm \sqrt{16+9} \\ & = \pm 5 \end{aligned}$

$T: y=x \pm 5$ will to cut the $E$ at $A\left(-\frac{16}{5}, \frac{9}{5}\right)$

$B\left(\frac{16}{5},-\frac{9}{5}\right)$

Also, $y=x$ will cut the $E$ at $C\left(\frac{12}{5}, \frac{12}{5}\right)$

$D\left(-\frac{12}{5},-\frac{12}{5}\right)$

$A B C D$ in not give in cyclic order

$\therefore$ it does not form any quadrilateral

$\therefore \quad$ No option should match

If order is not considered then

Area $=24$ sq. unit.

$\begin{aligned} & x^2+\frac{a^2 y^2}{b^2}=a^2 \\ & \Rightarrow y^2\left(1-\frac{a^2}{b^2}\right)=a^2\left(e^2-1\right)=a^2\left(1-\frac{b^2}{a^2}-1\right) \\ & =-b^2 \\ & \Rightarrow \frac{y^2\left(b^2-a^2\right)}{b^2}=-b^2 \Rightarrow y^2=\frac{b^4}{\left(a^2-b^2\right)} \\ & \text { Height }=|y|=\frac{b^2}{\sqrt{a^2-b^2}} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \text { Area }=(2 a e) \times \frac{1}{2} \times \frac{b^2}{\sqrt{a^2-b^2}}=30 \\ & =\frac{a b^2 e}{a \sqrt{1-\frac{b^2}{a^2}}}=b^2, a=\frac{17}{2} \end{aligned}\\ &\text { Distance between foci }=2 a e\\ &=17 \sqrt{1-\frac{b^2}{a^2}}=17 \sqrt{1-\frac{30 \times 4}{289}}=13 \end{aligned}$

To find the length of the latus rectum of the ellipse, we first recognize that the foci of the ellipse are given as $ F_1:(2,5) $ and $ F_2:(2,-3) $. This indicates that the major axis is aligned along the $ y $-axis.

Calculate the distance between the foci:

$ F_1F_2 = 8 $

The formula involving the distance between the foci and the eccentricity is:

$ F_1F_2 = 2be $

Here, the eccentricity $ e $ is given as $ \frac{4}{5} $. Thus, we can solve for $ b $:

$ 8 = 2b \cdot \frac{4}{5} \implies b = \frac{8}{2 \times \frac{4}{5}} = 5 $

Determine $ a^2 $:

Using the relationship between eccentricity, semi-minor axis $ b $, and semi-major axis $ a $:

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

Solving for $ a^2 $:

$ 1 - \frac{a^2}{25} = \frac{16}{25} \implies \frac{a^2}{25} = \frac{9}{25} \implies a^2 = 9 $

Thus, $ a = 3 $.

Compute the length of the latus rectum:

The formula for the length of the latus rectum $ L $ is:

$ L = \frac{2a^2}{b} $

Substituting the known values:

$ L = \frac{2 \times (9)}{5} = \frac{18}{5} $

Therefore, the length of the latus rectum of the ellipse is $ \frac{18}{5} $.

$\begin{aligned} &\text { Area }=\frac{1}{2} \times(a+b) \cdot 2 a=a(a+b)\\ &\text { Since, } e=\frac{1}{2} a e=2 \Rightarrow a=4, b=2 \sqrt{3}\\ &\Rightarrow \text { Area }=4(4+2 \sqrt{3})=8(2+\sqrt{3}) \end{aligned}$

$\begin{aligned} & \frac{x-\sqrt{5}}{\cos \theta}=\frac{y-\sqrt{5}}{\sin \theta}=r \\ & x=r \cos \theta+\sqrt{5} \\ & y=r \sin \theta+\sqrt{5} \\ & 25 x^2+36 y^2=900,(x, y) \text { lie on } E \\ & r^2\left[25 \cos ^2 \theta+36 \sin ^2 \theta\right]+r[50 \sqrt{5} \cos \theta+72 \sqrt{5} \sin \theta] \\ & \quad+25[5]+36[5]=900 \\ & r_1 r_2=\frac{900-305}{25+11 \sin ^2 \theta},\left(r_1 r_2\right)_{\max }=\frac{595}{25}=\frac{119}{5} \\ & \left|r_1 \cdot r_2\right|_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right) \text { at } \theta=0 \end{aligned}$

$\begin{aligned} & \frac{x^2}{36}+\frac{y^2}{25}=1 \\ & \text { At } y=\sqrt{5} \\ & \frac{x^2}{36}+\frac{1}{5}=1 \\ & \Rightarrow \frac{x^2}{36}=\frac{4}{5} \Rightarrow x= \pm \frac{12}{\sqrt{5}} \\ & P A^2+P B^2=(P A+P B)^2-2 P A \cdot P B=\left(\frac{24}{\sqrt{5}}\right)^2-2 \cdot \frac{119}{5} \\ & 5\left(P A^2+P B^2\right)=5\left(\frac{24^2}{5}-\frac{2.119}{5}\right)=24^2-238 \\ & =338 . \end{aligned}$

The length of the minor axis is equal to one-fourth of the distance between the foci. Mathematically, this can be expressed as:

$ 2b = \frac{1}{4}(2ae) $

This simplifies to:

$ b = \frac{ae}{4} $

Given that the relationship between $ b $, $ a $, and $ e $ is:

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

Substitute $ b = \frac{ae}{4} $ into the equation:

$ \left(\frac{ae}{4}\right)^2 = a^2 \cdot \left(1 - e^2\right) $

Expanding and simplifying gives:

$ \frac{a^2 e^2}{16} = a^2(1 - e^2) $

Divide both sides by $ a^2 $:

$ \frac{e^2}{16} = 1 - e^2 $

Rearrange and solve for $ e^2 $:

$ \frac{e^2}{16} + e^2 = 1 $

$ \frac{17e^2}{16} = 1 $

Solve for $ e $:

$ e^2 = \frac{16}{17} $

$ e = \frac{4}{\sqrt{17}} $

Thus, the eccentricity of the ellipse is:

$\frac{4}{\sqrt{17}}$

$\begin{aligned} & a=3 \sqrt{2}, b=3 \\ & \Rightarrow \quad e=\frac{1}{\sqrt{2}} \end{aligned}$

$\begin{aligned} & P S \cdot P S^{\prime}=2 a=6 \sqrt{2} \\ & \frac{P S+P S^{\prime}}{2} \geq \sqrt{P S \cdot P S^{\prime}} \\ & \Rightarrow\left(P S \times P S^{\prime}\right) \max =18 \end{aligned}$

Minima happens when $P$ lies on major axis

$\begin{aligned} & \Rightarrow \quad P=(3 \sqrt{2}, 0) \\ & P S=(3 \sqrt{2}-3) \cdot P S^{\prime}=(3 \sqrt{2}+3) \\ & \left(P S \cdot P S^{\prime}\right) \min =9 \\ & \left(P S \cdot P S^{\prime}\right) \min +\left(P S \cdot P S^{\prime}\right) \max =27 \end{aligned}$

Option (1)

$\begin{aligned} &\text { Equation of chord } \mathrm{T}=\mathrm{S}_1\\ &\begin{aligned} & \frac{5}{2}\left(\frac{\mathrm{x}}{9}\right)+\frac{1}{2}\left(\frac{\mathrm{y}}{4}\right)=\frac{25}{36}+\frac{1}{16} \\ & \Rightarrow \frac{5 \mathrm{x}}{18}+\frac{\mathrm{y}}{8}=\frac{100+9}{144}=\frac{109}{144} \\ & \Rightarrow 40 \mathrm{x}+18 \mathrm{y}=109 \\ & \Rightarrow \alpha=40, \beta=18 \\ & \Rightarrow \alpha+\beta=58 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & 2 \mathrm{ae}=4 \\ & 2 \mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4 \\ & \Rightarrow \mathrm{a}=2 \sqrt{3} \\ & \Rightarrow 1-\frac{\mathrm{b}^2}{12}=\frac{1}{3} \Rightarrow \mathrm{~b}^2=8 \\ & \text { Now } \frac{2 \mathrm{~b}^2}{\mathrm{a}} \cdot \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \\ & \Rightarrow \mathrm{~A}^2=2 \mathrm{~B} \\ & 1-\frac{\mathrm{A}^2}{\mathrm{~B}^2}=\frac{1}{3} \Rightarrow 1-\frac{2 \mathrm{~B}}{\mathrm{~B}^2}=\frac{1}{3} \Rightarrow \mathrm{~B}=3 \\ & \Rightarrow \mathrm{~A}^2=6 \\ & \frac{\mathrm{x}^2}{12}+\frac{\mathrm{y}^2}{8}=1 \ldots . .(1) \\ & \frac{\mathrm{x}^2}{6}+\frac{\mathrm{y}^2}{9}=1 \ldots . .(2) \end{aligned}\\ &\text { On solving (1) & (2) we get } \end{aligned}$

$\begin{aligned} &(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)\\ &\text { The four points are vertices of rectangle and its area }=\\ &\frac{24 \sqrt{6}}{5} \end{aligned}$

$\begin{aligned} &\text { If } \mathrm{m}\left(\sqrt{2}, \frac{4}{3}\right) \text { than equation of of } \mathrm{AB} \text { is }\\ &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4} \\ & \frac{\sqrt{2} \mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9} \end{aligned} \end{aligned}$

$\begin{aligned} &\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3} \text { put in ellipse }\\ &\begin{aligned} & \text { So, } \frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1 \\ & 4 x^2+36+2 x^2-12 \sqrt{2} x=36 \\ & 6 x^2-12 \sqrt{2} x=0 \\ & 6 x(x-2 \sqrt{2})=0 \\ & x=0 \& x=2 \sqrt{2} \\ & \text { So } y=2 \quad y=\frac{2}{3} \\ & \text { Length of chord }=\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2} \\ & \quad=\sqrt{8+\frac{16}{9}} \\ & \quad=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text { so } \alpha=22 \end{aligned} \end{aligned}$

$\begin{aligned} &\text { Equation of chord with given middle point }\\ &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \Rightarrow \frac{3 \mathrm{x}}{25}+\frac{\mathrm{y}}{16}-1=\frac{9}{25}+\frac{1}{16}-1 \\ & 48 \mathrm{x}+25 \mathrm{y}=144+25 \\ & 48 \mathrm{x}+25 \mathrm{y}=169 \text { Ans. } \end{aligned} \end{aligned}$

$\begin{aligned} & \text { Product of focal distances }=\left(\mathrm{a}+\mathrm{ex}_1\right)\left(\mathrm{a}-\mathrm{ex}_1\right) \\ & =\mathrm{a}^2-\mathrm{e}^2 \mathrm{x}_1^2=\mathrm{a}^2-\mathrm{e}^2(3) \\ & =\mathrm{a}^2-3 \mathrm{e}^2=\frac{7}{4} \Rightarrow \mathrm{a}^2=\frac{7}{4}+3 \mathrm{e}^2 \\ & \Rightarrow 4 \mathrm{a}^2=7+12 \mathrm{e}^2 \\ & \&\left(\sqrt{3}, \frac{1}{2}\right) \text { lines on } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\ & \therefore \frac{3}{\mathrm{a}^2}+\frac{1}{4 \mathrm{~b}^2}=1 \\ & \frac{3}{\mathrm{a}^2}+\frac{1}{4\left(\mathrm{a}^2\right)\left(1-\mathrm{e}^2\right)}=1 \\ & 12\left(1-\mathrm{e}^2\right)+1=4 \mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & 13-12 \mathrm{e}^2=\left(7+12 \mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \\ & \Rightarrow 13-12 \mathrm{e}^2=7-7 \mathrm{e}^2+12 \mathrm{e}^2-12 \mathrm{e}^4 \\ & \Rightarrow 12 \mathrm{e}^4-17 \mathrm{e}^2+6=0 \\ & \therefore \mathrm{e}^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \& \frac{2}{3} \\ & \therefore \mathrm{e}=\frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}} \\ & \therefore \text { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{T}=\mathrm{S}_1 \\ & \frac{\mathrm{x} \cdot 1}{4}+\frac{\mathrm{y} \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8} \\ & \mathrm{x}+\mathrm{y}=\frac{3}{2} \end{aligned}\\ &\text { solve with ellipse }\\ &\begin{aligned} \mathrm{PR} & =\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2} \\ & =\sqrt{2}\left|\mathrm{x}_2-\mathrm{x}_1\right| \end{aligned} \end{aligned}$

$\begin{aligned} & y_2=\frac{3}{2}-x_2 \\ & y_1=\frac{3}{2}-x_1 \\ & y_2-y_1=x_2-x_1 \\ & x^2+2 y^2=4 \\ & x^2+2\left(\frac{3}{2}-x\right)^2=4 \\ & 6 x^2-12 x+1=0 \\ & x_1+x_2=2 \\ & x_1 x_2=1 / 6 \\ & \left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2} \\ & \quad=\sqrt{4-4 / 6} \\ & P R=\sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15} \end{aligned}$

We are given an ellipse

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

and a hyperbola

$ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1. $

It is stated that “the distance between the foci of $E$ and the foci of $H$ is $2\sqrt{3}$.” A natural interpretation is that the foci of the ellipse are separated by

$ 2\sqrt{a^2-b^2}, $

and those of the hyperbola by

$ 2\sqrt{A^2+B^2}, $

and each distance is equal to $2\sqrt{3}$. That is, we have

$ 2\sqrt{a^2-b^2}=2\sqrt{3}\quad \Longrightarrow\quad a^2-b^2=3, $

and

$ 2\sqrt{A^2+B^2}=2\sqrt{3}\quad \Longrightarrow\quad A^2+B^2=3. $

We are also given that

$ a-A=2, $

and that the ratio of the eccentricities is

$ \frac{e_E}{e_H}=\frac{1}{3}, $

where the eccentricity of the ellipse is

$ e_E=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{3}}{a}, $

and the eccentricity of the hyperbola is

$ e_H=\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{3}}{A}. $

Thus the ratio becomes

$ \frac{e_E}{e_H}=\frac{\sqrt{3}/a}{\sqrt{3}/A}=\frac{A}{a}=\frac{1}{3}. $

This implies

$ a=3A. $

Now, using the condition $a-A=2$ together with $a=3A$, we get

$ 3A-A=2 \quad\Longrightarrow\quad 2A=2 \quad\Longrightarrow\quad A=1. $

Thus,

$ a=3. $

Next, for the ellipse we have

$ a^2-b^2=3 \quad \Longrightarrow\quad 9-b^2=3 \quad\Longrightarrow\quad b^2=6. $

For the hyperbola,

$ A^2+B^2=3 \quad \Longrightarrow\quad 1+B^2=3 \quad\Longrightarrow\quad B^2=2. $

The length of the latus rectum is given by the following formulas:

For the ellipse:

$ L_E=\frac{2b^2}{a}, $

For the hyperbola:

$ L_H=\frac{2B^2}{A}. $

Substitute the computed values:

For the ellipse:

$ L_E=\frac{2\times6}{3}=\frac{12}{3}=4, $

For the hyperbola:

$ L_H=\frac{2\times2}{1}=4. $

The sum of the lengths of the latus rectums is then

$ L_E+L_H=4+4=8. $

Thus, the answer is

$ 8. $

$\begin{aligned} & \mathrm{P} \equiv\left(3 \cos 30^{\circ}, 2 \sin 30^{\circ}\right) \equiv\left(\frac{3 \sqrt{3}}{2}, 1\right) \\ & \mathrm{Q} \equiv\left(3 \cos 60^{\circ}, 2 \sin 60^{\circ}\right) \equiv\left(\frac{3}{2}, \sqrt{3}\right) \\ & \mathrm{R}\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right), \mathrm{S}\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\end{aligned}$

Slope of $\mathrm{PQ}=\mathrm{m}_{\mathrm{PQ}}=\frac{\sqrt{3}-1}{\frac{3}{2}-\frac{3 \sqrt{3}}{2}}=-\frac{2}{3}$

Equation of line PQ

$ y-\sqrt{3}=-\frac{2}{3}\left(x-\frac{3}{2}\right) $

$\Rightarrow 2 x+3 y=3(\sqrt{3}+1) \quad$ option (A) is correct

Now

if $\mathrm{N}_2=\left(\mathrm{x}_2, 0\right)=\left(\frac{3}{2}, 0\right)$

$\left|\mathrm{N}_2 \mathrm{Q}\right|=\sqrt{3}$ and $\left|\mathrm{N}_2 \mathrm{~S}\right|=\frac{3 \sqrt{3}}{2}$

$\Rightarrow 3\left|\mathrm{~N}_2 \mathrm{Q}\right|=2\left|\mathrm{~N}_2 \mathrm{~S}\right| \quad$ option (C) is correct

Now, if $\mathrm{N}_1=\left(\mathrm{x}_1, 0\right) \Rightarrow \mathrm{N}_1=\left(\frac{3 \sqrt{3}}{2}, 0\right)$

$\Rightarrow\left|\mathrm{N}_1 \mathrm{P}\right|=1,\left|\mathrm{~N}_1 \mathrm{R}\right|=\frac{3}{2} \quad$ option (D) is incorrect

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 $F_2(-2,-5)$.

$x^2+4 y^2-4=0 \Rightarrow \frac{x^2}{4}+y^2=1$

Center $=(0,0), a=2, b=1$

The parametric equation of point on the ellipse

$ \begin{aligned} & x=a \cos \theta, y=b \sin \theta \\ & \text { at } \theta=\pi / 4 \\ & x_p=2 \times \frac{1}{\sqrt{2}}=\sqrt{2}, y_p=2 \times \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \end{aligned} $

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

Slope $=\frac{a^2 y_1}{b^2 x_1}=\frac{4 \times \frac{1}{\sqrt{2}}}{1 \times \sqrt{2}}=2$

$\therefore$ The equation of the normal at $p_{\text {at }}$

$ y-\frac{1}{\sqrt{2}}=2(x-\sqrt{2}) \Rightarrow y=2 x-\frac{3}{\sqrt2} $

and the normal intersects at point $Q(\alpha, \beta)$

$ \begin{aligned} & \therefore x^2+4\left(2 x-\frac{3}{\sqrt{2}}\right)^2=4 \\ & \Rightarrow x^2+16 x^2-24 \sqrt{2} x+18=4 \\ & \Rightarrow 17 x^2-24 \sqrt{2} x+14=0 \\ & \quad D=(-24 \sqrt{2})^2-4(17)(14)=200 \\ & \therefore \quad x=\frac{24 \sqrt{2} \pm \sqrt{200}}{34}=\frac{14 \sqrt{2}}{34}, \sqrt{2} \\ & \because \quad x=\sqrt{2} \text { corresponds to point } p \\ & \therefore \quad x=\frac{14 \sqrt{2}}{34}=\frac{7 \sqrt{2}}{17} \text { corresponds top } \\ & \text { So, } \alpha=\frac{7 \sqrt{2}}{17} \end{aligned} $