Ellipse

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 $=5+4=9$

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$

Let eccentricity of ellipse be ' $e^{\prime}$

Then $b^2=a^2\left(1-e^2\right)$

Here, $b^2=12, a^2=16$

$ \begin{aligned} \therefore \quad 12 & =16\left(1-e^2\right) \\ 1-e^2 & =\frac{3}{4} \end{aligned} $

or $\quad e^2=1-\frac{3}{4}=\frac{1}{4}$ or $e=\frac{1}{2}$

If $\alpha, \beta$ are the eccentric angles of the ends of a focal chord of the ellipse, then eccentricity is given by

$ e=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)} $

Here, $\alpha=\frac{\pi}{3}, \beta=\theta$, and $e=\frac{1}{2}$

$ \frac{1}{2}=\frac{\cos \left(\frac{\pi / 3-\theta}{2}\right)}{\cos \left(\frac{\pi / 3+\theta}{2}\right)} $

Multiplying in $N^r$ and $D^r$ by $2 \sin \left(\frac{\pi / 3+\theta}{2}\right)$

$ \frac{1}{2}=\frac{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3-\theta}{2}\right)}{2 \sin \left(\frac{\pi / 3+\theta}{2}\right) \cos \left(\frac{\pi / 3+\theta}{2}\right)} $

$ =\frac{\sin \left(\frac{\pi / 3+\theta+\pi / 3-\theta}{2}\right)+\sin \left(\frac{\pi / 3+\theta-\pi / 3+\theta}{2}\right)}{\sin (\pi / 3+\theta)} $

$ \begin{aligned} & \left\{\because 2 \sin A \cos B=\sin \left(\frac{A+B}{2}\right)+\sin \left(\frac{A-B}{2}\right)\right. \\ & \text { and } 2 \sin A \cos A=\sin 2 A\} \\ & \frac{1}{2}=\frac{\sin \frac{\pi}{3}+\sin \theta}{\sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta} \\ & \sin \frac{\pi}{3} \cos \theta+\cos \frac{\pi}{3} \sin \theta=2\left(\sin \frac{\pi}{3}+\sin \theta\right) \\ & \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=2\left(\frac{\sqrt{3}}{2}+\sin \theta\right) \end{aligned} $

$ \begin{aligned} & \text { } \frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta=\sqrt{3}+2 \sin \theta \\ & \qquad \begin{array}{c}or\,\,\, \frac{\sqrt{3}}{2} \cos \theta=\frac{3}{2} \sin \theta+\sqrt{3} \\ \cos \theta=\sqrt{3} \sin \theta+2 \end{array} \\ & \text { or } \frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta=1 \\ & \text { or } \cos \frac{\pi}{3} \cos \theta-\sin \frac{\pi}{3} \sin \theta=1 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { or } \cos \left(\frac{\pi}{3}+\theta\right)=1 \\ & \text { or } \cos \left(\frac{\pi}{3}+\theta\right)=\cos 0^{\circ} \end{aligned}\\ &\text { On comparing both sides, we get }\\ &\begin{array}{rlrl} & \frac{\pi}{3}+\theta =0^{\circ} \text { or } \theta=-\frac{\pi}{3} \\ \therefore & \tan \theta=\tan \left(\frac{-\pi}{3}\right) \Rightarrow \tan \theta=-\sqrt{3} \end{array} \end{aligned} $

Given ellipse

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

Here, $\quad a^2=1, b^2=2$

Equation of tangent

$ x+2 y+k=0 \text { or } 2 y=-x-k $

or       $ y=-\frac{x}{2}-\frac{k}{2} $

Here,    $ c=\frac{-k}{2}, m=-\frac{1}{2} $

From the condition of tangent

$ \begin{aligned} c & = \pm \sqrt{a^2 m^2+b^2} \\ \Rightarrow \frac{-k}{2} & = \pm \sqrt{(1)\left(\frac{1}{4}\right)+2} \\ \frac{-k}{2} & = \pm \sqrt{\frac{9}{4}} \Rightarrow \frac{-k}{2}= \pm \frac{3}{2} \end{aligned} $

or $k= \pm 3$

But $k>0$, so, $k=3$

So, equation of tangent, $y=-\frac{x}{2}-\frac{3}{2}$ and point is $\left(\frac{1}{\sqrt{2}}, 1\right)$

The equation of the normal to the ellipse at point ( $x_1, y_1$ ) is given by

$ \begin{aligned} & \frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2 \\ \Rightarrow \quad & \frac{1 \times x}{1 / \sqrt{2}}-\frac{2 \times y}{1}=1-2 \\ & \sqrt{2} x-2 y=-1 \text { or } \sqrt{2} x-2 y+1=0 \end{aligned} $

$a \alpha^2+b \beta^2+c \alpha \beta+d=0$ is the transformed equation of

$ 4 x^2+\sqrt{3} x y+5 y^2-4=0 $

where,

$ \begin{align*} & \alpha=\frac{\sqrt{3}}{2} x+\frac{y}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \beta=\frac{-x}{2}+\frac{\sqrt{3} y}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

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

$ \begin{aligned} x & =\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2} \\ \Rightarrow \quad y & =\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta \end{aligned} $

Putting the value of $x$ and $y$ in

$ 4 x^2+\sqrt{3} x y+5 y^2-4=0, $

$ \begin{aligned} &\text { we get }\\ &\begin{aligned} & 4\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)^2 +\sqrt{3}\left(\frac{\sqrt{3}}{2} \alpha+\frac{\beta}{2}\right)\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right) +5\left(\frac{\alpha}{2}-\frac{\sqrt{3}}{2} \beta\right)^2-4=0 \end{aligned} \end{aligned} $

Given ellipse,

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

Point $P(\sqrt{2} \cos \theta, \sin \theta)$ lie in ellipse Tangent at $P(\sqrt{2} \cos \theta, \sin \theta)$ on ellipse is,

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

Intercept on line is, $A\left(\frac{\sqrt{2}}{\cos \theta}, 0\right)$ and $ B\left(0, \frac{1}{\sin \theta}\right) $

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

$ \begin{array}{rlrl} \therefore & h =\frac{\sqrt{2}}{2 \cos \theta} \text { and } k=\frac{1}{2 \sin \theta} \\ \Rightarrow & \cos \theta =\frac{1}{\sqrt{2} h} \text { and } \sin \theta=\frac{1}{2 k} \end{array} $

Squaring and adding, we get

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

∴ Locus is, $\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$

Given ellipse, $\frac{x^2}{16}+\frac{y^2}{12}=1$

End point of latus rectum

$ =\left( \pm a e, \frac{2 b^2}{a}\right)=(2,3) $

Equation of tangent at $(2,3)$ of ellipse

$ \frac{2 x}{16}+\frac{3 y}{12}=1 \Rightarrow \frac{x}{8}+\frac{y}{4}=1 $

Area of quadrilateral

$ \begin{aligned} A B C D & =4 \times \frac{1}{2} \times O A \times O B \\ & =4 \times \frac{1}{2} \times 8 \times 4=64 \end{aligned} $

Since the foci of given ellipse lie on the $Y$-axis, it is vertical ellipse.

Let the required equation be $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$,

where $a^2>b^2$.

Foci are $(0, \pm c)=(0, \pm 1) \Rightarrow c=1$

$a=$ length of the semi-major axis

$ \begin{aligned} & =\frac{1}{2} \times \sqrt{5}=\frac{\sqrt{5}}{2} \\ \therefore \quad a^2 & =\frac{5}{4} \end{aligned} $

Now, $c^2=a^2-b^2 \Rightarrow 1=\left(\frac{\sqrt{5}}{2}\right)^2-b^2$

$ \Rightarrow \quad b^2=\frac{5}{4}-1=\frac{1}{4} $

∴ The required equation is

$ \begin{aligned} & \frac{x^2}{1 / 4}+\frac{y^2}{5 / 4}=1 \Rightarrow 4 x^2+\frac{4 y^2}{5}=1 \\ \Rightarrow & 20 x^2+4 y^2=5 \end{aligned} $

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

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

Length of major axis is the diameter of circle

$ \begin{array}{rlrl} x^2+y^2 =18 \\ \therefore 2 a=2(3 \sqrt{2})^2 \Rightarrow a & =3 \sqrt{2} \\ e^2 & =1-\frac{b^2}{a^2} \\ \Rightarrow & \frac{8}{9} =1-\frac{b^2}{18} \Rightarrow b^2=2 \end{array} $

∴ Equation of ellipse is $\frac{x^2}{18}+\frac{y^2}{2}=1$

Equation of circle is $x^2+y^2=18$

Let $(h, k)$ be the pole, then equation of polar with respect to ellipse is

$ \frac{x h}{18}+\frac{y k}{2}=1 $

Since this is the tangent of circle

$ \begin{aligned} \therefore 3 \sqrt{2}=\left|\frac{1}{\sqrt{\frac{h^2}{(18)^2}+\frac{k^2}{4}}}\right| \\ \frac{h^2}{18^2}+\frac{k^2}{4}= & \frac{1}{18} \end{aligned} $

$ \begin{array}{r} \frac{h^2}{18}+\frac{9 k^2}{2}=1 \\ \therefore \text { Locus is } \frac{x^2}{18}+\frac{9 y^2}{2}=1 \end{array} $

Let the equation of ellipse be,

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

It passes through $(3 \sqrt{2}, \sqrt{10})$

$ \therefore \quad \frac{18}{a^2}+\frac{10}{b^2}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, foci are ( $\pm 4,0$ )

$ \begin{array}{ll} \therefore & a e=4 \Rightarrow a^2 e^2=16 \\ \Rightarrow & a^2\left(1-\frac{b^2}{a^2}\right)=16 \Rightarrow a^2-b^2=16 \\ \Rightarrow & b^2=a^2-16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

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

$ \begin{aligned} & \frac{18}{a^2}+\frac{10}{a^2-16}=1 \\ \Rightarrow & 18 a^2-288+10 a^2=a^4-16 a^2 \\ \Rightarrow & a^4-44 a^2+288=0 \\ \Rightarrow & \left(a^2-36\right)\left(a^2-8\right)=0 \end{aligned} $

$ \begin{aligned} &\begin{array}{lll} \Rightarrow & a^2=36,8 & \\ \Rightarrow & a^2=36 & {\left[\because a^2 \neq 8\right. \text { as }} \\ & & \left.b^2=a^2-16\right] \end{array}\\ &\text { Now, } \quad a^2 e^2=16\\ &e^2=\frac{16}{a^2}=\frac{16}{36} \Rightarrow e=\frac{4}{6}=\frac{2}{3} \end{aligned} $

We know that, the product of the length of the perpendicular drawn from the foci to the tangent of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $b^2$.

Also, length $=9 \quad \therefore b^2=9$

Equation of tangent

$ \begin{aligned} y & =\frac{-3}{4} x+3 \sqrt{2} \\ \therefore \quad 3 \sqrt{2} & = \pm \sqrt{a^2\left(\frac{-3}{4}\right)^2+b^2} \\ 18 & =\frac{9}{16} a^2+9 \Rightarrow 2-1=\frac{a^2}{16} \\ \Rightarrow \quad a^2 & =16 \\ \therefore \quad e & =\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4} \end{aligned} $

Given equation of ellipse

$16{x^2} + 11{y^2} = 256 \Rightarrow {{{x^2}} \over {16}} + {{11{y^2}} \over {256}} = 1$

Equation of tangent of ellipse at $\left( {4\cos \phi ,{{16} \over {\sqrt {11} }}\sin \phi } \right)$ is

${{4x\cos \phi } \over {16}} + {{11 \times 16y} \over {256\sqrt {11} }}\sin \phi = 1$

$ \Rightarrow {{x\cos \phi } \over 4} + {{\sqrt {11} } \over {16}}\sin \phi y = 1$

is also a tangent of circle ${x^2} + {y^2} - 2x = 15$

$ \Rightarrow {(x - 1)^2} + {y^2} = 16$

radius of circle = 4 and centre = (1, 0)

Now, ${{x\cos \phi } \over 4} + {{\sqrt {11} } \over {16}}\sin \phi y = 1$ is tangent of circle.

$\therefore$ $4 = \left| {{{{{\cos \phi } \over 4} - 1} \over {\sqrt {{{{{\cos }^2}\phi } \over {16}} + {{11} \over {256}}{{\sin }^2}\phi } }}} \right| \Rightarrow \phi = \pm {\pi \over 3}$

$\therefore$ $\phi = {\pi \over 3}$