An ellipse intersects the hyperbola 2 x^2-2 y | vedclass

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Question

Given, $2 x^2-2 y^2=1$

$\Rightarrow \quad \frac{x^2}{\left(\frac{1}{2}ight)}-\frac{y^2}{\left(\frac{1}{2}ight)}=1 \quad \ldots 	ext { (i) }$

Vedclass · Accepted Answer

$(A,B)$

Vedclass · Answer

Given, $2 x^2-2 y^2=1$

$\Rightarrow \quad \frac{x^2}{\left(\frac{1}{2}ight)}-\frac{y^2}{\left(\frac{1}{2}ight)}=1 \quad \ldots 	ext { (i) }$

$	ext { Eccentricity of hyperbola }=\sqrt{2}$

$	ext { So eccentricity of ellipse }=1 / \sqrt{2}$

Eccentricity of hyperbola $=\sqrt{2}$

So eccentricity of ellipse $=1 / \sqrt{2}$

Let equation of ellipse be

$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a > b)$

$	herefore $$ \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow $$ \frac{b^2}{a^2}=\frac{1}{2} \Rightarrow a^2=2 b^2$

$	herefore $$ x^2+2 y^2=2 b^2 \ldots$

Let ellipse and hyperbola intersect at

$A\left(\frac{1}{\sqrt{2}} \sec 	heta, \frac{1}{\sqrt{2}} 	an 	hetaight)$

On differentiating Eq. ($i$),

$4 x-4 y \frac{d y}{d x} $$ =0$

$\frac{d y}{d x} $$ =\frac{x}{y}$

$\left.\frac{d y}{d x}ight|_{	ext {at } A} $$ =\frac{\sec 	heta}{	an 	heta}=\operatorname{cosec} 	heta$

$\Rightarrow \quad \frac{d y}{d x}=\frac{x}{y}$

and differentiating Eq. ($ii$),

$2 x+4 y \frac{d y}{d x}=0$

$\left.\frac{d y}{d x}ight|_{	ext {at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} 	heta$

Since, ellipse and hyperbola are orthogonal.

$	herefore $$ -\frac{1}{2} \operatorname{cosec}^2 	heta=-1$

$\Rightarrow $$ \operatorname{cosec}^2 	heta=2$

$\Rightarrow $$ 	heta= \pm \frac{\pi}{4}$

$	herefore $$ A\left(1, \frac{1}{\sqrt{2}}ight) 	ext { or }\left(1,-\frac{1}{\sqrt{2}}ight)$

$	herefore$ Form Eq. (i),

$1+2\left(\frac{1}{\sqrt{2}}ight)^2=2 b^2 \Rightarrow b^2=1$

Equation of ellipse is $x^2+2 y^2=2$

Coordinate of foci $( \pm a e, 0)=\left( \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0ight)=( \pm 1,0)$

Hence, option $(a)$ and $(b)$ are correct.

If major axis is along $y$-axis, then

$\frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}} \Rightarrow b^2=2 a^2 $

$	herefore $$ 2 x^2+y^2=2 a^2 $

$\Rightarrow \quad y^{\prime}=-\frac{2 x}{y} \Rightarrow $$ y^{\prime}\left(\frac{1}{\sqrt{2}} \sec 	heta, \frac{1}{\sqrt{2}} 	an 	hetaight)=\frac{-2}{\sin 	heta}$

As ellipse and hyperbola are arthogonal

$	herefore $$ -\frac{2}{\sin 	heta} \cdot \operatorname{cosec} 	heta=-1 $

$\Rightarrow $$ \operatorname{cosec}^2 	heta=1 \Rightarrow 	heta= \pm \frac{\pi}{4} $

$	herefore $$ 2 x^2+y^2=2 a^2 $

$\Rightarrow $$ 2+\frac{1}{2}=2 a^2 $

$\Rightarrow $$ a^2=\frac{5}{4}$

$	herefore $$ 2 x^2+y^2=\frac{5}{2}, 	ext { corresponding foci are }(0, \pm 1) .$

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