Ellipse

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

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} $

$ \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 $=2 \alpha \beta$

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 $

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 $

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

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

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

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

Since, $b>a$.

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

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

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

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

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

Thus, focal distance

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

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

Thus, required focal distances are 4, 2.

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

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

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

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

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

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

Ellipse : $x^2+3 y^2=6$

$\Rightarrow \quad \frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1$

Let the eccentric angle of the point be $\theta$, then its coordinates are $(\sqrt{6} \cos \theta, \sqrt{2} \sin \theta)$.

Since, the distance of the point from the centre is 2 units.

Therefore, $(\sqrt{6} \cos \theta-0)^2+(\sqrt{2} \sin \theta-0)^2=2^2$

$\begin{array}{cc} \Rightarrow & 6 \cos ^2 \theta+2\left(1-\cos ^2 \theta\right)=4 \\ \Rightarrow & 4 \cos ^2 \theta=2 \\ \Rightarrow & \cos \theta= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow & \theta=\frac{\pi}{4}, \frac{3 \pi}{4} \end{array}$

Given that, $AP+BP=2a$

Using distance formula,

$AP=\sqrt{(x+ae)^2+y^2}$

and $B P=\sqrt{(x-a e)^2+y^2}$

Now, $\sqrt{(x+a e)^2+y^2}+\sqrt{(x-a e)^2+y^2}=2 a$

Squaring both sides,

$\begin{gathered} (x+a e)^2+y^2+(x-a e)^2+y^2 \\ +2 \sqrt{(x+a e)^2+y^2} \sqrt{(x-a e)^2+y^2}=4 a^2 \\ \Rightarrow \quad 2 x^2+2 y^2+2 a^2 e^2-4 a^2 \\ =-2 \sqrt{(x+a e)^2+y^2} \sqrt{(x-a e)^2+y^2} \\ x^2+y^2+a^2 e^2-2 a^2=-\sqrt{(x+a e)^2+y^2} \\ \\ \cdot \sqrt{(x-a e)^2+y^2} \end{gathered}$

Squaring both sides,

$\begin{aligned} & a^4 e^4-4 a^4 e^2+4 a^4+2 a^2 x^2 e^2-4 a^2 x^2 \\ & +2 a^2 y^2 e^2-4 a^2 y^2+x^4+2 x^2 y^2+y^4 \\ & =a^4 e^4-2 a^2 x^2 e^2+2 a^2 y^2 e^2+x^4+2 x^2 y^2+y^4 \\ & \Rightarrow \quad 4\left(a^4-a^4 e^2-a^2 x^2-a^2 y^2+a^2 x^2 e^2\right)=0 \\ & \Rightarrow \quad a^2\left(a^2-a^2 e^2-x^2-y^2+x^2 e^2\right)=0 \\ & \Rightarrow a^2\left(1-e^2\right)-x^2-y^2+x^2 e^2=0 \\ \end{aligned}$

$\begin{array}{left}\Rightarrow \quad b^2-x^2-y^2+x^2\left(1-\frac{b^2}{a^2}\right)=0 \quad\left[\because b^2=a^2\left(1-e^2\right)\right] \\ \Rightarrow \quad x^2\left[1-\frac{b^2}{a^2}-1\right]-y^2+b^2=0 \\ \Rightarrow x^2\left(-\frac{b^2}{a^2}\right)-y^2=-b^2 \\ \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\left(\text { divide by } b^2\right)\end{array}$

Let two points be $A\left(a \cos \theta_1 b \sin \theta_1\right)$ and $\left(a \cos \theta_2, b \sin \theta_2\right)$.

$\begin{aligned} & O: \operatorname{origin}(0,0) \\ & \qquad \begin{aligned} m_{O A} & =\frac{b \sin \theta_1}{a \cos \theta_1}=\frac{b}{a} \tan \theta_1 \\ m_{O B} & =\frac{b}{a} \tan \theta_2 \\ \Rightarrow \quad m_{O A} \times m_{O B} & =-1 \Rightarrow \frac{b^2}{a^2} \tan \theta_1 \tan \theta_2=-1 \end{aligned} \end{aligned}$

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

Distance between foci $=2 a e=6$

$\begin{aligned} & \Rightarrow \quad a e=3 \\ & \text { and length of minor axis }=2 b=8 \\ & \Rightarrow \quad b=4 \\ & \because \quad(a e)^2=a^2-b^2 \\ & \Rightarrow \quad 3^2=a^2-4^2 \Rightarrow a^2=25 \\ & \text { Now, } e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5} \\ \end{aligned}$

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

Length of semi-major axis $=5$

Length of semi-minor axis $=4$

Centre $C=(0,0)$

If $P(x, y)$ be any point on the ellipse, then

maximum value of $C P=$ Length of the semi-major axis $=5$

minimum value of $C P=$ Length of semi-minor axis $=4$

Their sum $=5+4=9$