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} $

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

$\mathrm{r}=4$

$\because$ 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. $

$\begin{aligned} & g(10)=10 \\ & a=f(g(10))=f(10)=109 \\ & f(3)=18 \\ & b=g(f(3))=g(18)=2 \\ & \frac{x^2}{109}+\frac{y^2}{2}=1 \\ & e=\sqrt{1-\frac{2}{109}}=\sqrt{\frac{107}{109}} \\ & I=\frac{2 b^2}{a}=\frac{2 \times 2}{\sqrt{109}} \end{aligned}$

$ \begin{aligned} 8 e^2+l^2 & =\frac{8 \times 107}{109}+\frac{16}{109} \\ & =8 \end{aligned} $

Equation of circle with $A B$ as diameter

$\begin{aligned} & \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\ & \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0 \end{aligned}$

Comparing, $k=6$

Latus rectum of ellipse

$\begin{aligned} & x^2+9 y^2=k^2=6^2 \\ & \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \\ & \text { L.R }=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3} \\ & m=4 \\ & n=3 \\ & 2 m+n=8+3=11 \end{aligned}$

$\begin{aligned} & \mathrm{h}=3 \cos \theta \\\\ & \mathrm{k}=\frac{18}{7} \sin \theta\end{aligned}$

$\begin{aligned} & \therefore \text { locus }=\frac{\mathrm{x}^2}{9}+\frac{49 \mathrm{y}^2}{324}=1 \\\\ & \mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\end{aligned}$

Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a > b$, the eccentricity $ e $ is given by the formula:

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

It is provided that the eccentricity $ e $ is $ \frac{1}{\sqrt{2}} $ (given), so we can equate the two expressions for eccentricity:

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

Squaring both sides to eliminate the square root gives:

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

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

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

Taking the square root on both sides:

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

$ a = b\sqrt{2} $

Now, for the ellipse, the length of the latus rectum is given by the formula:

$ \text{Length of Latus Rectum (L)} = \frac{2b^2}{a} $

It's provided that the length of the latus rectum $ L $ is $ \sqrt{14} $, so substitute the known values to find $ b $:

$ \sqrt{14} = \frac{2b^2}{b\sqrt{2}} = \frac{2b}{\sqrt{2}} $

$ b\sqrt{2} = \sqrt{14} $

$ b^2 = \frac{14}{2} $

$ b^2 = 7 $

And since $ a = b\sqrt{2} $, we can find $ a^2 $:

$ a^2 = (b\sqrt{2})^2 $

$ a^2 = 7 \cdot 2 $

$ a^2 = 14 $

Now we have an ellipse with $ a^2 = 14 $ and $ b^2 = 7 $. The equation of a hyperbola similar to the given ellipse but with the terms subtracted is:

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

For the hyperbola, the square of the eccentricity $ e' $ is given by:

$ (e')^2 = 1 + \frac{b^2}{a^2} $

Substitute the values we've found for $ a^2 $ and $ b^2 $ into the formula for the square of the hyperbola's eccentricity:

$ (e')^2 = 1 + \frac{b^2}{a^2} $

$ (e')^2 = 1 + \frac{7}{14} $

$ (e')^2 = 1 + \frac{1}{2} $

$ (e')^2 = \frac{3}{2} $

Therefore, the square of the eccentricity of the hyperbola is $ \frac{3}{2} $, which corresponds to option C.

$\begin{aligned} & \text { Slope of axis }=\frac{1}{2} \\ & y-3=\frac{1}{2}(x-2) \\ & \Rightarrow 2 y-6=x-2 \\ & \Rightarrow 2 y-x-4=0 \\ & 2 x+y-6=0 \\ & 4 x+2 y-12=0 \\ & \alpha+1.6=4 \Rightarrow \alpha=2.4 \\ & \beta+2.8=6 \Rightarrow \beta=3.2 \end{aligned}$

Ellipse passes through $(2.4,3.2)$

$\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{\mathrm{a}^2}+\frac{\left(\frac{32}{10}\right)^2}{\mathrm{~b}^2}=1$ ..... (1)

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

$\Rightarrow a^2=2 b^2$

Put in (1) $\Rightarrow b^2=\frac{328}{25}$

$\Rightarrow\left(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)^2=\frac{4 \mathrm{~b}^2}{\mathrm{a}^2} \times \mathrm{b}^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25} $

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

$\begin{aligned} & \text { ae }=3 \\ & \frac{\mathrm{a}}{\mathrm{e}}=\frac{25}{3} \\ & \mathrm{a}=5 \\ & \mathrm{~b}=4 \end{aligned}$

Length of $L R=\frac{2 b^2}{a}=\frac{32}{5}$

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

Equation of chord with given middle point.

$\begin{aligned} & T=S_1 \\ & \frac{x}{25}+\frac{y}{40}=\frac{1}{25}+\frac{1}{100} \\ & \frac{8 x+5 y}{200}=\frac{8+2}{200} \\ & y=\frac{10-8 x}{5} \quad \text{.... (i)} \end{aligned}$

$\frac{x^2}{25}+\frac{(10-8 x)^2}{400}=1$ (put in original equation)

$\begin{aligned} & \frac{16 x^2+100+64 x^2-160 x}{400}=1 \\ & 4 x^2-8 x-15=0 \\ & x=\frac{8 \pm \sqrt{304}}{8} \\ & x_1=\frac{8+\sqrt{304}}{8} ; x_2=\frac{8-\sqrt{304}}{8} \end{aligned}$

Similarly, $y=\frac{10-18 \pm \sqrt{304}}{5}=\frac{2 \pm \sqrt{304}}{5}$

$\begin{aligned} & \mathrm{y}_1=\frac{2-\sqrt{304}}{5} ; \mathrm{y}_2=\frac{2+\sqrt{304}}{5} \\ & \text { Distance }=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2} \\ & =\sqrt{\frac{4 \times 304}{64}+\frac{4 \times 304}{25}}=\frac{\sqrt{1691}}{5} \\ \end{aligned}$

$ \begin{aligned} & \frac{x^2}{36}+\frac{y^2}{4}=1 \\\\ & T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\\\ & T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\\\ & N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}} \end{aligned} $

$ \begin{aligned} & \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\ & 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\\\ & N: 3 x-\sqrt{3} y=8 \sqrt{3} \\\\ & A(0,4) \quad B(0,-8) \\\\ & \text { C: } x^2+(y-4)(y+8)=0 \\\\ & \text { Line } x=2 \sqrt{5} \\\\ & 20+y^2+4 y-32=0 \\\\ & y^2+4 y-12=0 \\\\ & (y+6)(y-2)=0 \end{aligned} $

$ \begin{aligned} & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & C: x^2+y^2+4 y-32=0 \\\\ & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & T: x x_1+y y_1+2 y+2 y_1-32=0 \\\\ & T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\\\ & \quad 2 \sqrt{5} x-4 y=44 \\\\ & T_1: \sqrt{5} x-2 y=22 .......(i) \\\\ & T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\\\ & 2 \sqrt{5} x+4 y=28 \\\\ & T_2: \sqrt{5} x+2 y=14 .......(ii) \end{aligned} $

From (i) and (ii), we get

$ \begin{aligned} & \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\\\ & \alpha^2-\beta^2=\frac{304}{5} \end{aligned} $

Given, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\frac{x^2}{4} + \frac{y^2}{9} = 1$. This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$-axis and semi-minor axis $b=2$ along the $x$-axis.

OP is the distance from origin O to point P, which is given by :

$OP =r_1 = \sqrt{\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2} = \sqrt{\frac{12}{7} + \frac{36}{7}} = \sqrt{\frac{48}{7}} = 2\sqrt{\frac{12}{7}}.$

1. Let's represent the given point $P\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$ in polar coordinates. We can write $P$ as $(r_1 \cos \theta, r_1 \sin \theta)$. Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1 \cos \theta$ and $y=r_1 \sin \theta$ into the equation of the ellipse gives us :

$\frac{r_1^2 \cos ^2 \theta}{4}+\frac{r_1^2 \sin ^2 \theta}{9}=1$

Simplifying this, we obtain :

$\frac{\cos ^2 \theta}{4}+\frac{\sin ^2 \theta}{9}=\frac{7}{48} \quad \text{--- (equation 1)}$

2. Similarly, if we represent the point $R$ as $(-r_2 \sin \theta, r_2 \cos \theta)$, (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\pi/2$ radians. In terms of sin and cos, $\sin(\theta + \pi/2) = \cos(\theta)$ and $\cos(\theta + \pi/2) = -\sin(\theta)$.) it too should satisfy the equation of the ellipse. We have :

$\frac{r_2^2 \sin ^2 \theta}{4}+\frac{r_2^2 \cos ^2 \theta}{9}=1$

Simplifying this, we obtain :

$\frac{\sin ^2 \theta}{4}+\frac{\cos ^2 \theta}{9}=\frac{1}{r_2^2} \quad \text{--- (equation 2)}$

3. From equations (1) and (2), we have :

$\frac{1}{r_2^2}=\frac{1}{4}+\frac{1}{9}-\frac{7}{48}=\frac{31}{144}$

4. Now, note that lines $PQ$ and $RS$ are perpendicular and pass through the origin, so $PQ = 2OP$ and $RS = 2OR$. Thus,

$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4} \left(\frac{1}{r_1^2} + \frac{1}{r_2^2} \right)$

5. Substituting the values of $r_1$ and $r_2$, we obtain :

$\frac{1}{PQ^2} + \frac{1}{RS^2} = \frac{1}{4}\left(\frac{7}{48}+\frac{31}{144}\right)=\frac{13}{144} = \frac{p}{q}$

6. Hence, $p = 13$ and $q = 144$.

7. So, the final answer $p+q = 13 + 144 = 157$.

The given ellipse has the equation :

$x^2+4y^2=36$

We can rewrite this as :

$\frac{x^2}{6^2} + \frac{y^2}{(6/2)^2} = 1$

This shows that it is an ellipse centered at (0,0) with semi-major axis a = 6 along the x-axis and semi-minor axis b = 3 along the y-axis.

The equation of a circle with center (2,0) and radius r is :

$(x-2)^2 + y^2 = r^2$

Substituting y^2 from the ellipse equation into the circle equation gives us :

$x^2 - 4x + 4 + \frac{36 - x^2}{4} = r^2$

Solving this equation leads to :

$3x^2 - 16x + 52 - 4r^2 = 0$

For the roots of this quadratic equation to be real (which they must be, since they represent real intersection points), the discriminant (D) must be greater than or equal to zero :

$D = b^2 - 4ac = (-16)^2 - 4\times3\times(52 - 4r^2) = 256 - 12\times52 + 48r^2$

Setting D = 0 gives the minimum value for r (the radius of the inscribed circle) :

$256 - 624 + 48r^2 = 0$

$48r^2 = 368$

$r^2 = \frac{368}{48} = \frac{23}{3}$

And we're asked for the value of 12r², so :

$12r^2 = 12 \times \frac{23}{3} = 92$

So, the correct answer is Option B : 92.

We have, $E_K=K x^2+K^2 y^2=1, K=1,2, \ldots 20$

$\Rightarrow \frac{x^2}{\frac{1}{K}}+\frac{y^2}{\frac{1}{K^2}}=1$



Equation of $A B$ is $\frac{x}{\frac{1}{\sqrt{K}}}+\frac{y}{\frac{1}{K}}=1$

or $ \sqrt{K} x+K y=1$

$r_K$ is the radius of circle $C_K$,

So, $r_K=$ perpendicular distance from $(0,0)$ to $\mathrm{AB}$

$ \begin{aligned} r_K & =\frac{|0+0-1|}{\sqrt{(\sqrt{K})^2+K^2}} \\\\ & =\frac{1}{\sqrt{K+K^2}} \\\\ \frac{1}{r_K^2} & =K+K^2 \end{aligned} $

$ \begin{aligned} \sum_{K=1}^{20} \frac{1}{r_K^2} & =\sum_{K=1}^{20} K+\sum_{K=1}^{20} K^2 \\\\ & =\frac{20 \times 21}{2}+\frac{20 \times 21 \times 41}{6} \\\\ & =210+2870=3080 \end{aligned} $

We have, equation of ellipse : $15 x^2+19 y^2=285$

or $ \frac{x^2}{19}+\frac{y^2}{15}=1$

Let the coordinate of center of circle be $(0,0)$.

Equation of circle is $x^2+y^2=16$

Equation of tangent of ellipse is

$ \begin{gathered} y=m x \pm \sqrt{19 m^2+15} \text { or } \\\\ m x-y \pm \sqrt{19 m^2+15}=0 \end{gathered} $

It is also tangent to the circle $x^2+y^2=16$

Perpendicular distance from center of circle to tangent $=4$

$ \frac{\left|0-0 \pm \sqrt{19 m^2+15}\right|}{\sqrt{m^2+1}}=4 $

On squaring both side, we get

$ \begin{gathered} 19 m^2+15=16 m^2+16 \\\\ 3 m^2=1 \\\\ m^2=\frac{1}{3} \\\\ m= \pm \frac{1}{\sqrt{3}} \end{gathered} $

$ \tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6} \text { with } X \text {-axis or } $$\frac{\pi}{3}$ with $Y$-axis

Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi}{3}$.

The given equation of the ellipse is

$ \begin{aligned} & x^2+9 y^2=9 ~..........(i)\\\\ & \Rightarrow \frac{x^2}{9}+\frac{y^2}{1}=1 \end{aligned} $

Now, equation of line $A B$ is

$ x+3 y=3 ~$...........(ii)



Now, point of intersection of line $x+3 y=3$ and circle

$ \begin{array}{lr} &x^2+y^2=9 \\\\ &\therefore (3-3 y)^2+y^2=9 \\\\ &\Rightarrow 9-18 y+9 y^2+y^2=9 \\\\ &\Rightarrow -18 y+10 y^2=0 \\\\ &\Rightarrow 18 y=10 y^2 \\\\ &\Rightarrow y=0, \frac{9}{5} \end{array} $

Now, from figure, we clearly see that vertices $A, P$ and $O$ makes a $\triangle A P O$,

whose area is $\frac{1}{2} \times$ Base $\times$ Height

$ \begin{aligned} & \text { Area }=\frac{1}{2} \times 3 \times \frac{9}{5}=\frac{27}{10}=\frac{m}{n} ~~~~[Given]\\\\ & \therefore m=27, n=10 \\\\ & \Rightarrow m-n=27-10=17 \end{aligned} $

Let $E$ be the person speak, English

$\therefore n(E)=75$

and $H$ be the person speak Hindi

$\therefore n(H)=40$

Let number of persons who speak both English and Hindi are $t$.


$\begin{array}{rlrl} &\therefore \alpha+t+\beta =100 ........(i) \\\\ & \alpha+t =75........(i) \\\\ &\text { and }\beta+t =40 ........(iii)\end{array}$

From Equations (i) and (ii), $\beta=25$

From Equations (i) and (iii), $\alpha=60$ and from Eq. (i), $t=15$

We have, equation of ellipse

$ \begin{aligned} 25\left(\beta^2 x^2+\alpha^2 y^2\right) & =\alpha^2 \beta^2 \\\\ \Rightarrow 25\left(\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}\right) & =1 \end{aligned} $

$\begin{aligned} & \Rightarrow 25\left(\frac{x^2}{3600}+\frac{y^2}{625}\right)=1 \\\\ & \Rightarrow \frac{x^2}{144}+\frac{y^2}{25}=1 \\\\ & \therefore \text { Eccentricity, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{25}{144}}=\frac{\sqrt{119}}{12}\end{aligned}$

Equation of normal is

$2 x \sec \theta-b y \operatorname{cosec} \theta=4-b^{2}$

Distance from $(0,0)=\frac{4-b^{2}}{\sqrt{4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta}}$

Distance is maximum if

$4 \sec ^{2} \theta+b^{2} \operatorname{cosec}^{2} \theta$ is minimum

$\Rightarrow \tan ^{2} \theta=\frac{\mathrm{b}}{2}$

$\Rightarrow \frac{4-b^{2}}{\sqrt{4 \cdot \frac{b+2}{2}+b^{2} \cdot \frac{b+2}{b}}}=1$

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

$\Rightarrow(b+2)(b-1)=0$

$\Rightarrow b=1$

$\therefore e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

Equation of tangent having slope m.

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

which passes through $(1,3)$

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

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

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

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

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

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

Acute angle between the tangents is given by

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

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

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

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

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

So, $a = 7$

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

So, $b = 2\sqrt 6 $

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

$e = {5 \over 7}$

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

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

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

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

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

Also, given hyperbola,

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

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

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

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

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

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

Also given,

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

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

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

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

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

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

Slope of tangent to y = e^x curve

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

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

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

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

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

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

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

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

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

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

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

$\therefore$ $a = \sqrt 2 $

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

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

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

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

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

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

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

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

From (i)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$ \Rightarrow a = 4$

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

The point of intersection of the curves ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ and ${x^2} + {y^2} = 3$ in the first quadrant is $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$

Now slope of tangent to the ellipse ${{{x^2}} \over 9} + {{{y^2}} \over 1} = 1$ at $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$ is

${m_1} = - {1 \over {3\sqrt 3 }}$

And slope of tangent to the circle at $\left( {{3 \over 2},{{\sqrt 3 } \over 2}} \right)$ is m₂ $ = - \sqrt 3 $

So, if angle between both curves is $\theta$ then

$\tan \theta = \left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - {1 \over {3\sqrt 3 }} + \sqrt 3 } \over {1 + \left( { - {1 \over {3\sqrt 3 }}\left( { - \sqrt 3 } \right)} \right)}}} \right|$

$ = {2 \over {\sqrt 3 }}$

Option (b)