Hyperbola

We have,

${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$

${{{{(x - \sqrt 2 )}^2}} \over 4} - {{{{(y + \sqrt 2 )}^2}} \over 2} = 1$

$a = 2,b = \sqrt 2 $

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

Now, $e = \sqrt {{{{a^2} + {b^2}} \over {{a^2}}}} = \sqrt {{3 \over 2}} $

Area, $ = {1 \over 2}$ $\times$ Base $\times$ Height

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

$ = {1 \over 2}{{(\sqrt 3 - \sqrt 2 ) \times 2} \over {\sqrt 2 }}$

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

$ = \left( {\sqrt {{3 \over 2}} - 1} \right)$

Equation of normal at $P\left( {x,y} \right)$ is $Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$

Coordinate of $G$ at $X$ axis is $\left( {X,0} \right)$ (let)

$\therefore$ $0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$

$ \Rightarrow X = x + y{{dy} \over {dx}}$ $\therefore$ Co-ordinate of $G\left( {x + y{{dy} \over {dx}},0} \right)$

Given distance of $G$ from origin $=$ twice of the abscissa of $P.$

as distance cannot be $-ve,$ therefore abscissa $x$ should be $+ve$

$\therefore$ $x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$

On Integrating $ \Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$

$\therefore$ the curve is a hyperbola

Given, equation of hyperbola is ${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$

We know that the equation of hyperbola is

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ Here, ${a^2} = {\cos ^2}\alpha $ and ${b^2} = {\sin ^2}\alpha $

We know that, ${b^2} = {a^2}\left( {{e^2} - 1} \right)$

$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$

$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$

$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $

$\therefore$ $ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$

Co-ordinate of foci are $\left( { \pm \alpha e,0} \right)\,\,$ i.e. $\left( { \pm 1,0} \right)$

Hence, abscissae of foci remain constant when $\alpha $ varies.

The length of transverse axis $ = 2\sin \theta = 2a$

$a = \sin \theta $

also for ellipse $3{x^2} + 4{y^2} = 12$

or ${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$

${a^2} = 4,{b^3} = 3$

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

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

$\therefore$ Focus of ellipse $ = \left( {2 \times {1 \over 2},0} \right) \Rightarrow (1,0)$

As hyperbola is confocal with ellipse, focus of hyperbola is = (1, 0)

$ae = 1 \Rightarrow \sin \theta \times e = 1$

$e = \cos ec\theta $

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

${\sin ^2}\theta (\cos e{c^2}\theta - 1) = {\cos ^2}\theta $

$\therefore$ equation of hyperbola is

${{{x^2}} \over {{{\sin }^2}\theta }} - {{{y^2}} \over {{{\cos }^2}\theta }} = 1$

${x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$

The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.

Here, the numbers 25 and 16 are the squares of the lengths of the axes. This means the semi-major axis $ a = 5 $ and the semi-minor axis $ b = 4 $.

Step 1: Find the eccentricity of the ellipse.

The eccentricity of an ellipse is given by the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$

Let’s plug in the values: $ b^2 = 16 $, $ a^2 = 25 $

So, $ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $

Step 2: Find the focus of the ellipse.

The foci are at $ (\pm \sqrt{a^2 - b^2}, 0) $.

So, $ a^2 - b^2 = 25 - 16 = 9 $. Take square root: $ \sqrt{9} = 3 $.

This means the foci are at $ (\pm 3, 0) $.

Step 3: Eccentricity of the hyperbola.

The problem tells us: Product of eccentricities is 1.

The ellipse's eccentricity $ e_{ellipse} = \frac{3}{5} $. Let $ e_{hyperbola} $ be the hyperbola's eccentricity.

So, $ e_{hyperbola} \times \frac{3}{5} = 1 $. $ e_{hyperbola} = \frac{1}{\frac{3}{5}} = \frac{5}{3} $

Step 4: Hyperbola passes through the focus (3, 0).

Since the axes of the hyperbola match the ellipse, and it passes through (3, 0), write: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Plug in $ x = 3 $, $ y = 0 $:

$\frac{3^2}{a^2} - 0 = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9$

So the hyperbola equation is: $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$

Step 5: Use the eccentricity formula for the hyperbola.

For a hyperbola, eccentricity $ e_{hyperbola} = \sqrt{1 + \frac{b^2}{a^2}} $

We know $ e_{hyperbola} = \frac{5}{3} $, $ a^2 = 9 $. Set up the equation:

$\sqrt{1 + \frac{b^2}{9}} = \frac{5}{3}$

Square both sides: $1 + \frac{b^2}{9} = \frac{25}{9}$

So, $ b^2 = 16 $.

Step 6: Final equation of the hyperbola.

So, the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$ with $ a^2 = 9, b^2 = 16 $.

Step 7: Find the foci of the hyperbola.

The foci are at $ (\pm \sqrt{a^2 + b^2}, 0) $.

Plug in $ a^2 = 9, b^2 = 16 $: $ \sqrt{9 + 16} = \sqrt{25} = 5 $.

So, the foci are at $ (\pm 5, 0) $.

Tangent to the hyperbola ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ is

$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $

Given that $y = \alpha x + \beta $ is the tangent of hyperbola

$ \Rightarrow m = \alpha $ and ${a^2}{m^2} - {b^2} = {\beta ^2}$

$\therefore$ ${a^2}{\alpha ^2} - {b^2} = {\beta ^2}$

Locus is ${a^2}{x^2} - {y^2} = {b^2}$ which is hyperbola.

Let, any point on the hyperbola is $(3 \sec \theta, 2 \tan \theta)$

$\therefore$ Chord of contact of the circle

$x^{2}+y^{2}=9$ w.r.t $(3 \sec \theta, 2 \tan \theta)$ is as follows

$(3 \sec \theta) x+(2 \tan \theta) y=9$ ..... (i)

let $\left(x_{1}, y_{1}\right)$ be the mid point of the chord of the contact

$x x_{1}+y y_{1}=x_{1}^{2}+y_{1}^{2}$ ..... (ii)

From (i) and (ii), we have

$\begin{aligned} \frac{3 \sec \theta}{x_{1}} & =\frac{2 \tan \theta}{y_{1}}=\frac{9}{x_{1}^{2}+y_{1}^{2}} \\ \Rightarrow \quad \sec \theta & =\frac{9 x_{1}}{3\left(x_{1}^{2}+y_{1}^{2}\right)}\quad ....(\mathrm{iii}) \\ \text { and } \tan \theta & =\frac{9 y_{1}}{2\left(x_{1}^{2}+y_{1}^{2}\right)}\quad ....(\mathrm{iv})\end{aligned}$

Now, by eliminating $\theta$ from (iii) and (iv), we get

$\frac{81 x_{1}^{2}}{9\left(x_{1}^{2}+y_{1}^{2}\right)^{2}}-\frac{81 y_{1}^{2}}{4\left(x_{1}^{2}+y_{1}^{2}\right)^{2}}=1$

$\left(\because \sec ^{2} \theta-\tan ^{2} \theta=1\right)$

Required locus $\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{\left(x^{2}+y^{2}\right)^{2}}{81}$

${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$

$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$

$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$

$\therefore$ Foci $ = \left( { \pm 3,0} \right)$

$\therefore$ foci of ellipse $=$ foci of hyperbola

$\therefore$ for ellipse $ae=3$ but $a=4,$

$\therefore$ $e = {3 \over 4}$

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

$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$