Ellipse

$ \begin{aligned} &\\ &\begin{aligned} & \text { Focus }=(1,-2) \\ & \text { directrix }=3 x+4 y-15=0 \\ & \therefore \quad \text { Length of latusrectum }=4 \\ & \frac{2 b^2}{a}=4 \Rightarrow b^2=2 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∵ Distance between focus and directrix }\\ &\begin{aligned} & \frac{a}{e}-a e \\ & \quad \frac{a}{e}-a e=\left|\frac{3-8-15}{5}\right| \\ & \Rightarrow \quad a\left(\frac{1}{e}-e\right)=4 \\ & \quad\left[\because e^2=1-\frac{b^2}{a^2} \Rightarrow 1-e^2=\frac{b^2}{a^2}\right] \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{a}{e}\left(1-e^2\right)=4 \\ \therefore & \frac{a}{e} \times \frac{b^2}{a^2}=4 \\ \Rightarrow & \frac{2 a}{e \times a}=4 \Rightarrow e=\frac{1}{2} \end{array} $

Reason is correct explanation of Assertion.

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

Equation of tangent having slope $\frac{1}{3}$ is

$ y=m n \pm \sqrt{a^2 m^2+b c} $

$ \Rightarrow \quad y=\frac{x}{3} \pm \sqrt{\frac{a^2}{9}+b^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

∴ Equation (i) is normal of circle

$ (x+1)^2+(y+1)^2=1 $

∴ It passes through $(-1,-1)$

$ \begin{array}{ll} & -1=\frac{-1}{3} \pm \sqrt{\frac{a^2}{9}+b^2} \\ \Rightarrow & \left(\frac{2}{3}\right)^2=\frac{a^2}{9}+b^2 \Rightarrow 9 b^2=4-a^2 \\ \because & a>b \Rightarrow a^2>b^2 \\ \Rightarrow & a^2>\frac{4-a^2}{9} \Rightarrow 9 a^2>4-a^2 \\ \Rightarrow & 10 a^2>4 \Rightarrow a^2>\frac{4}{10} \\ \Rightarrow & a^2>\frac{2}{5} \text { and } 4-a^2>0 \\ & a^2<4 \\ \therefore& a^2 \in\left(\frac{2}{5}, 4\right) \end{array} $

$9 x^2+4 y^2=144 \Rightarrow \frac{x^2}{16}+\frac{y^2}{36}=1$

Equation of tangent at $(4 \cos \theta, 6 \sin \theta)$

$ \frac{x \cos \theta}{4}+\frac{y \sin \theta}{6}=1 $

$ \therefore A\left(\frac{4}{\cos \theta}, 0\right), B\left(0, \frac{6}{\sin \theta}\right) $

Area of $\triangle O A D=\frac{1}{2} \times \frac{4}{\cos \theta} \cdot \frac{6}{\sin \theta}$

$ =\frac{24}{\sin 2 \theta} $

∴ Minimum area $=24$

$ \begin{array}{ll} \because & \theta=\frac{\pi}{4} \\ \therefore & \alpha=\frac{4}{\sqrt{2}}, \beta=\frac{6}{\sqrt{2}} \Rightarrow \alpha \beta=12 \end{array} $

∴ Minimum area $=2 \alpha \beta$

Given, equation: $9 x^2+4 y^2=72$

Differentiate w.r.t $x$, we get

$ \begin{aligned} & 18 x+8 y \cdot \frac{d y}{d x}=0 \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{-18 x}{8 y}=\frac{-9 x}{4 y} \end{aligned} $

Slope of tangent at point $(2,3)$ is

$ m_t=\frac{-9(2)}{4(3)}=\frac{-18}{12}=\frac{-3}{2} $

So, equation of tangent using point-slope form is

$ \begin{aligned} & & (y-3) & =-\frac{3}{2}(x-2) \\ &\Rightarrow & 2 y-6 & =-3 x+6 \\ & \Rightarrow & 3 x+2 y & =12 \end{aligned} $

Now, slope of normal, $m_n=\frac{-1}{m_t}=\frac{2}{3}$

Now, slope of normal, $m_n=\frac{-1}{m_t}=\frac{2}{3}$

So, equation of normal is

$ \begin{aligned} & & (y-3) & =\frac{2}{3}(x-2) \\ & \Rightarrow & 3 y-9 & =2 x-4 \\ & \Rightarrow & 2 x-3 y & =-5 \end{aligned} $

Now, for the $x$-intercept of the tangent line, put $y=0$, then

$ \begin{aligned} & & 3 x+2(0) & =12 \\ \Rightarrow & & x & =4 \end{aligned} $

so $(4,0)$

For the $x$-intercept of the normal line, put $y=0$, then

$ \begin{aligned} & 2 x-3(0)=-5 \\ \Rightarrow \quad & x=-\frac{5}{2}, \text { so, }\left(\frac{-5}{2}, 0\right) \end{aligned} $

Now, base of triangle $=$ Distance $\mathrm{b} / \mathrm{w} x$-intercepts of the tangent and normal lines

$ b=4-\left(\frac{-5}{2}\right)=\frac{13}{2} $

And, height of triangles is the $y$-coordinate of the point $(2,3)$ is $h=3$

So, area $=\frac{1}{2} b h=\frac{1}{2}\left(\frac{13}{2}\right) \times 3=\frac{39}{4}$

$\therefore$ Area $=\frac{39}{4}$ sq. units

$S(x, y) \equiv x^2+2 y^2-2 x+8 y+5$

Thus, $\frac{\partial s}{\partial x}=2 x-2 \frac{\partial s}{\partial y}=4 y+8$

at $P \equiv(\sqrt{2}+1,-1)$

$ 2 x-2=2(\sqrt{2}+1)-2=2 \sqrt{2} $

and $4 y+8=-4+8=4$

So, $\mathbf{n}=\langle 2 \sqrt{2}, 4\rangle$

So, slope $=\frac{4}{2 \sqrt{2}}=\sqrt{2}$

Then, equation of normal (at $\sqrt{2}+1,-1$ )

$ \begin{aligned} & y-(-1)=\sqrt{2}(x-(\sqrt{2}+1)) \\ \Rightarrow & y+1=\sqrt{2} x-2-\sqrt{2} \\ \Rightarrow & y-\sqrt{2} x+\sqrt{2}+3=0 \\ \Rightarrow & \sqrt{2} x-y-3-\sqrt{2}=0 \\ \Rightarrow & \sqrt{2} x-y=3+\sqrt{2} \end{aligned} $

$ \begin{aligned} & \text { } 4 x^2+9 y^2-16 x+54 y+61=0 \\ & \Rightarrow \quad(2 x-4)^2+(3 y+9)^2=36 \\ & \Rightarrow \quad \frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1 \end{aligned} $

Which is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

where $X=x-2 Y=y+3, a=3$ and $b=2$

As we know equation of director circle (because locus of two perpendicular pair of tangents shows its director circle) is $x^2+y^2=a^2+b^2$

$ \begin{aligned} & \Rightarrow \quad(x-2)^2+(y+3)^2=(3)^2+(2)^2 \\ & \Rightarrow \quad x^2+4-4 x+y^2+9+6 y=13 \\ & \Rightarrow \quad x^2+y^2-4 x+6 y=0 \end{aligned} $

Area of ellipse $A_1=\pi a b$

Let $\left(x_1, y_1\right)$ be a point on the ellipse Let $(a e, 0)$ be the focus Let $(h, k)$ be the mid-point.

$ \begin{aligned} & h=\frac{x_1+a e}{2}, k=\frac{y_1}{2} \\ & x_1=2 h-a e, y_1=2 k \\ \Rightarrow \quad & \frac{(2 h-a e)^2}{a^2}+\frac{(2 k)^2}{b^2}=1 \\ \Rightarrow \quad & \frac{\left(h-\frac{a e}{2}\right)^2}{\left(\frac{a}{2}\right)^2}+\frac{k^2}{\left(\frac{b}{2}\right)^2}=1 \end{aligned} $

$\Rightarrow$ Semi-major axis $=\frac{a}{2}$

Semi-minor axis $=\frac{b}{2}$

The area of the locus ellipse

$ \begin{aligned} & A_2=\pi \cdot \frac{a}{2} \cdot \frac{b}{2}=\frac{a b \pi}{4} \\ \Rightarrow \quad & \frac{A_1}{A_2}=\frac{\pi a b}{\frac{\pi a b}{4}}=\frac{4}{1} \\ \Rightarrow \quad & A_1: A_2=4: 1 \end{aligned} $

Given equation of ellipse

$ \begin{aligned} & 4 x^2+9 y^2-36=0 \\ & \Rightarrow \quad \frac{x^2}{9}+\frac{y^2}{4}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Equation of pair of tangent from P( - 3, 2)

$ \begin{aligned} S S_1 & =T^2 \\ \left(\frac{x^2}{9}+\frac{y^2}{4}-1\right) & \left(\frac{(-3)^2}{9}+\frac{(2)^2}{4}-1\right) \\ & =\left(-\frac{3(x)}{9}+\frac{2(y)}{4}-1\right)^2 \\ \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}-1 & =\left(-\frac{x}{3}+\frac{y}{2}-1\right)^2 \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}-1=\frac{x^2}{9}+\frac{y^2}{4}-\frac{x y}{3}-y+\frac{2 x}{3} \\ & \Rightarrow \frac{x y}{3}-\frac{2 x}{3}+y-2=0 \\ & \Rightarrow x y-2 x+3 y-6=0 \\ & \Rightarrow(x+3)(y-2)=0 \\ & \quad \quad x+3=0 \text { and } y-2=0 \\ & \Rightarrow x=-3 \text { and } y=2 \end{aligned} $

The tangent lines are $x=-3$ (a vertical line)

and $y=2$ (a horizontal line)

$\therefore$ These ar perpendicular

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

We have, $2 x^2+y^2=1$

And $x-y=-1$

Now make homogeneous equation of given ellipse.

$ \begin{aligned} & 2 x^2+y^2=(y-x)^2 \\ \Rightarrow \quad & 2 x^2+y^2=y^2+x^2-2 x y \\ \Rightarrow \quad & x^2+2 x y+0 \cdot y^2=0 \end{aligned} $

Angle $A O B$

$ \begin{array}{ll} & \tan \theta=\frac{2 \sqrt{1-0}}{1}=2 \\ \therefore \quad & \theta=\tan ^{-1} 2 \end{array} $

We have a circle $x^2+y^2=\frac{25}{4}$ and Ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$

Let equation of tangent to the ellipse be

$ y=m x \pm \sqrt{9 m^2+4} $

Given this tangent also touches the circle

$\because$ Condition of tendency

$ \frac{\sqrt{9 m^2+4}}{\sqrt{1+m^2}}=\frac{5}{2} $

On squaring both sides, we get

$ \begin{aligned} & 4\left(9 m^2+4\right)=25\left(1+m^2\right) \\ \Rightarrow \quad & 36 m^2+16=25+25 m^2 \\ \Rightarrow \quad & 11 m^2=9 \\ \therefore \quad & m^2=\frac{9}{11} \end{aligned} $

Given, ellipse is $x^2+2 y^2=2$

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

Now, equation of tangent in parametric form is

$ \frac{x}{\sqrt{2}} \cos \theta+y \sin \theta=1 $

Now, coordinate of $X$-intercept

$ =(\sqrt{2} \sec \theta, 0) A $

Coordinate of $Y$-intercept

$ =(0, \operatorname{cosec} \theta) B $

Let $P(h, k)$ be the points which is the mid-points of $A$ and $B$.

$ \begin{aligned} h & =\frac{\sqrt{2} \sec \theta}{2} \text { and } k=\frac{\operatorname{cosec} \theta}{2} \\ \sec \theta & =\sqrt{2} h \text { and } \operatorname{cosec} \theta=2 k \\ \cos \theta & =\frac{1}{\sqrt{2 h}} \text { and } \sin \theta=\frac{1}{2 k} \end{aligned} $

Using $\sin ^2 \theta+\cos ^2 \theta=1$

$ \frac{1}{2 h^2}+\frac{1}{4 k^2}=1 $

Taking locus of $P(h, k)$, we get

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

Let $\left(x_1, y_1\right)$ be the point of intersection of tangents to the ellipse, where it is intersected by the line $l x+m y+n=0$. Then, the equation of the chord of contact of tangents is

$ \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Clearly, $l x+m y+n=0$ and (i) represents the same line.

$ \begin{array}{ll} \therefore & \frac{x_1}{a^2 l}=\frac{y_1}{b^2 m}=\frac{1}{-n} \\ \Rightarrow & x_1=\frac{-a^2 l}{n}, y_1=\frac{-b^2 m}{n} \end{array} $

Thus, the required point is $\left(\frac{-a^2 l}{n}, \frac{-b^2 m}{n}\right)$

Let the rectangle have vertices at $( \pm x, \pm y)$ on the ellipse. From the ellipse equation

$ x=a \cos \theta, y=b \sin \theta $

The area of the rectangle is

$ A=4 x y=4(a \cos \theta \cdot b \sin \theta) $

Using the identity $\sin 2 \theta=2 \sin \theta \cos \theta$, we rewrite $A=2 a b \sin 2 \theta$

To maximize $A$, we maximize $\sin 2 \theta$, which reaches its maximum value of 1 when $2 \theta=\frac{\pi}{2}$.

Thus, the maximum area is

$ A_{\max }=2 a b $

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

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

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

$\operatorname{Ar}(\Delta \mathrm{ORT})=\frac{3}{2}$

$\begin{aligned} & \left|\frac{1}{2} \times 3 \times 2 \sin \theta\right|=\frac{3}{2} \\\\ & \sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{11 \pi}{6} \\\\ & \mathrm{~T}\left(\frac{3 \sqrt{3}}{2},-1\right)\end{aligned}$

Tanget at $(0,2) $,

$\frac{x(0)}{9}+\frac{y(2)}{4}=1 \Rightarrow y=2$ .........(1)

Tangent at $\left(\frac{3 \sqrt{3}}{2},-1\right) $,

$\frac{x\left(\frac{3 \sqrt{3}}{2}\right)}{9}+\frac{y(-1)}{4}=1$ ...........(2)

$\therefore$ By solving (1) & (2) $\Rightarrow \mathrm{p}=3 \sqrt{3}, \mathrm{q}=2$

$\Rightarrow$ Option $(\mathrm{A})$ is Correct.

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} $

We have, equation of ellipse is

$ \frac{x^2}{4}+\frac{y^2}{6}=1 $

Equation of tangent at point $(\sqrt{2}, \sqrt{3})$ is

$ \begin{array}{r} \frac{\sqrt{2} x}{4}+\frac{\sqrt{3}}{6}=1 \\ 3 \sqrt{2} x+2 \sqrt{3} y=12 \end{array} $

$ \Rightarrow \quad 18 x^2+12 y^2+12 \sqrt{6} x y=144 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Slope of tangent $T_1$

$ =\frac{-3 \sqrt{2}}{2 \sqrt{3}}=-\sqrt{\frac{3}{2}} $

Slope of tangent $T_2=\sqrt{\frac{2}{3}}$

Equation of tangent $T_r$ is

$ \begin{aligned} & y=\sqrt{\frac{2}{3}} x \pm \sqrt{4 \times \frac{2}{3}+6} \\ & \sqrt{3} y-\sqrt{2} x= \pm \sqrt{26} \\ & 3 y^2+2 x^2-2 \sqrt{6} x y=26 \end{aligned} $

or $\quad 18 y^2+12 x^2-12 \sqrt{6} x y=156\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ \begin{aligned} 30\left(x^2+y^2\right) & =300 \\ x^2+y^2 & =10 \end{aligned} $

As $(\alpha, \beta)$ are the intersecting point of $T_1$ and $T_2$,

$ \alpha^2+\beta^2=10 $

To find the length of the latus rectum of the given ellipse, start with the equation:

$ 16x^2 + 25y^2 = 400 $

First, rewrite it in standard form. Divide through by 400:

$ \frac{x^2}{400/16} + \frac{y^2}{400/25} = 1 $

This simplifies to:

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

Here, $ a^2 = 25 $ and $ b^2 = 16 $. For an ellipse in this form, the length of the latus rectum $ L $ is given by:

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

Calculate $ L $ using the values of $ a $ and $ b $:

$ L = \frac{2 \times 16}{\sqrt{25}} = \frac{32}{5} $

Therefore, the length of the latus rectum is $\frac{32}{5}$.

We have, equation of ellipse

$ \frac{x^2}{9}+\frac{y^2}{25}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

from above equation, we obtained

$ a=3, b=5 \text { i.e. } b^2>a^2 $

Here, product of perpendicular from focus is equal to length of semi-minor axis.

$ \Rightarrow \quad\left(S^{\prime} f^{\prime}\right) \times(S f)=\left(a^2\right)=(3)^2=9 $

The product of perpendicular from the two foci of the ellipse on the tangent at any point is 9 .

Given the ellipse equation:

$ x^2 + 4y^2 - 4 = 0 $

This can be rearranged into the standard form of an ellipse:

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

Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we identify that $a^2 = 4$ and $b^2 = 1$, so $a = 2$ and $b = 1$.

Area of the Ellipse ($A_1$):

The area of an ellipse is given by the formula $A = \pi a b$.

Thus,

$ A_1 = \pi \times 2 \times 1 = 2\pi $

Director Circle:

The equation of the director circle of an ellipse is $x^2 + y^2 = a^2 + b^2$.

Here,

$ a^2 + b^2 = 4 + 1 = 5 $

The equation of the director circle becomes:

$ x^2 + y^2 = 5 $

The radius $r$ of this director circle is $\sqrt{5}$. So, the area ($A_2$) is:

$ A_2 = \pi r^2 = \pi \times 5 = 5\pi $

Auxiliary Circle:

The auxiliary circle has the same center as the ellipse and its radius is equal to the semi-major axis of the ellipse, $a$. Hence, the equation is:

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

The radius $r$ is $\sqrt{4} = 2$. Thus, the area ($A_3$) is:

$ A_3 = \pi \times 4 = 4\pi $

Finally, calculate $A_2 + A_3 - A_1$:

$ A_2 + A_3 - A_1 = 5\pi + 4\pi - 2\pi = 7\pi $

Therefore, the result is $7\pi$.

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

Equation of chord with given mid-point $\left(x_1, y_1\right)$ is

$ \frac{x x_1}{4}+\frac{y y_1}{9}=\frac{x_1^2}{4}+\frac{y_1^2}{9} $

Since, ( 1,1 ) is mid-points, we get

$ \begin{aligned} & \\ & \frac{x}{4}+\frac{y}{9}=\frac{1}{4}+\frac{1}{9} \\ \Rightarrow & 9 x+4 y=13 \\ \Rightarrow & x+\frac{4}{9} y=\frac{13}{9} \end{aligned} $

But $\quad x+\alpha y=\beta \quad \text { [given] } $

On comparing, we get

$ \alpha=\frac{4}{9}, \beta=\frac{13}{9} $

So, $\quad \alpha+1=\frac{4}{9}+1$

$ =\frac{4+9}{9}=\frac{13}{9}=\beta $

$\therefore \quad \alpha+1=\beta$

Here, $a=2$

Coordinate of $F$ and $F^{\prime}$ are ( $2 e, 0$ ), ( $-2 a, 0$ ).

Distance between $F F^{\prime}=4 e$

$ \begin{array}{rlrl} \text { Area of } \triangle F B F^{\prime} & =\frac{1}{2} \times b \times 4 e \\ \sqrt{3} & =2 b e \\ b e & =\frac{\sqrt{3}}{2} \text { or } b=\frac{3}{2 e} \\ \Rightarrow & e & =\sqrt{1-\frac{b^2}{4}} \\ \Rightarrow \quad & e^2 & =1-\frac{3}{4 e^2 \times 4} \end{array} $

$ \begin{array}{ll} \Rightarrow \quad & 16 e^4-16 e^2+3=0 \\ \Rightarrow \quad & 16 e^4-12 e^2-4 e^2+3=0 \\ & \left(4 e^2-1\right)\left(4 e^2-3\right)=0 \\ & e^2=\frac{1}{4} \text { or } e^2=\frac{3}{4} \\ & e=\frac{1}{2} \text { or } \frac{\sqrt{3}}{2}[\because e \text { can't be negatiec }] \end{array} $

To solve this problem, we start by considering the given ellipse equation:

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

The equation of a tangent to this ellipse with a slope of 2 is:

$ y = 2x \pm \sqrt{4a^2 + b^2} $

Rearranging, the equation becomes:

$ 2x - y \pm \sqrt{4a^2 + b^2} = 0 \quad \text{(equation i)} $

This tangent also touches a circle with center $(0,0)$ and radius 2, given by the equation:

$ x^2 + y^2 = 4 $

For the tangent line to touch the circle, the perpendicular distance from the center of the circle to the line should equal the radius of 2. The formula for the perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $

Applying this to equation (i), with $A = 2, B = -1, C = \pm \sqrt{4a^2 + b^2}$, and the center $(0,0)$, we get:

$ \frac{|\pm \sqrt{4a^2 + b^2}|}{\sqrt{4 + 1}} = 2 $

Solving this gives:

$ \frac{\sqrt{4a^2 + b^2}}{\sqrt{5}} = 2 $

Squaring both sides, we obtain:

$ 4a^2 + b^2 = 20 \quad \text{(equation ii)} $

Now, applying the AM-GM inequality to find the maximum value of $ab$, we have:

$ \frac{4a^2 + b^2}{2} \geq \sqrt{4a^2 \cdot b^2} $

Substituting from (ii):

$ \frac{20}{2} \geq \sqrt{4a^2b^2} $

Simplifying:

$ 10 \geq 2ab \quad \Rightarrow \quad ab \leq 5 $

Therefore, the maximum value of $ab$ is 5.

We have, equation of ellipse is

$ \begin{aligned} & 3 x^2+8 y^2=K \\ & \Rightarrow \quad \frac{x^2}{\frac{K}{3}}+\frac{y^2}{\frac{K}{8}}=1 \end{aligned} $

Here, $a^2=\frac{K}{3}$ and $b^2=\frac{K}{8}$

Equation of normal $=4 x-3 y-5=0$

Or $y=\frac{4}{3} x-\frac{5}{3}$

Here, $m=\frac{4}{3}$ and

$\begin{aligned} \frac{\left(a^2-b^2\right) m}{\sqrt{a^2+b^2 m^2}} & =\frac{5}{3} \\ \frac{\left(\frac{K}{3}-\frac{K}{8}\right) \times \frac{4}{3}}{\sqrt{\frac{K}{3}+\frac{K}{8} \times \frac{16}{9}}} & =\frac{5}{3} \\ K^2 & =20 K \\ K & =20 \quad[\because K \neq 0]\end{aligned}$

Since, $(-2, m)$ lies cilipse

$ \because 3(4)+8 m^2=20 \Rightarrow m= \pm 1 $

Equation of tangent of ellipse at $(-2,1)$ is

$ -6 x+8 y=20 \Rightarrow 3 x-4 y+10=0 $

$ \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

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

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

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

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

$\Rightarrow b=1$

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