Question

An ellipse intersects the hyperbola $2x^2-2y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal to th a t of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

• (a) equation of ellipse is $x^2+2y^2=2$
• (b) the foci of ellipse are ($\pm$1, 0)
• (c) equation of ellipse is $x^2+2y^2=4$
• (d) the foci of ellipse are $(\pm\sqrt {2},0)$

Answer 1

The Correct Option is B

Given,2x^2-2y^2=1
$\Rightarrow \frac{x^2}{\bigg(\frac{1}{2}\bigg)}-\frac{y^2}{\bigg(\frac{1}{2}\bigg)}=1 \, \, \, \, \, \, \, \, \, \, \, \, (i) $
Eccentricity of hyperbola $=\sqrt 2$
So, eccentricity of ellipse $=\frac{1}{\sqrt 2}$
Let equation of ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, [where a>b]$
$\therefore\, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{1}{\sqrt 2}=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{b^2}{a^2}=\frac{1}{2} \Rightarrow \, \, a^2=2b^2$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, x^2+2y^2=2b^2 \, \, \, \, \, \, \, \, \, \, \, ........(ii)$
Let ellipse and hyperbola intersect at
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, A\bigg(\frac{1}{\sqrt 2} sec\theta,\frac{1}{\sqrt 2 tan \theta}\bigg)$
On differentiating E (i), we get
$\, \, \, \, \, \, \, \, \, \, 4x-4y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{x}{y}$
$ \, \, \, \, \, \, \, \, \, \, \, \, \frac{dy}{dx}|_at A=\frac{sec\theta}{tan\theta}=cosec\theta$
and on differentiating E (ii), we get
$2x+4y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}|_at A=\frac{sec\theta}{tan\theta}=cosec\theta$
Since, ellipse and hyperbola are orthogonal.
$\therefore -\frac{1}{2}cosec^2\theta=-1$
$\Rightarrow\, \, \, \, \, \, \, \, \, \, -\frac{1}{2}cosec^2\theta=2\Rightarrow\theta=\pm\frac{\pi}{4}$
$\, \, \, \, \, \, \, \, A\bigg(1,\frac{1}{\sqrt2}\bigg) or \bigg(1,-\frac{1}{\sqrt2}\bigg)$
Form E (i), $1+2\bigg(\frac{1}{\sqrt2}\bigg)^2=2b^2$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, b^2=1$
Equation of ellipse is $x^2+2y^2=2$
Coordinates of foci ($\pm$ ae, 0)$=\bigg(\pm\sqrt2.\frac{1}{\sqrt2},0)=(\pm1,0)$
If major axis is along Y-axis, then
$\, \, \, \, \, \, \, \, \, \, \, \frac{1}{\sqrt2}=\sqrt{1-\frac{a^2}{b^2}}\Rightarrow b^2=2a^2$
$\therefore \, \, \, \, \, \, \, \, \, 2x^2+y^2=2a^2 \Rightarrow Y'=-\frac{2x}{y}$
$\Rightarrow \, \, \, \, \, \, \, \, y'\bigg(\frac{1}{\sqrt2}sec\theta,\frac{1}{\sqrt2}tan\theta)=\frac{-2}{sin\theta}$
As ellipse and hyperbola are orthogonal
$\therefore \, \, \, \, \, \, \, \, -\frac{2}{sin\theta}.cosec\theta=-1$
$\Rightarrow \, \, \, \, \, \, \, \, \, cosec^2\theta=\frac{1}{2}\Rightarrow \theta=\pm\frac{\pi}{4}$
$\therefore\, \, \, \, \, \, \, \, 2x^2+y^2=2a^2$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, 2+\frac{1}{2}=2a^2\Rightarrow $
$\therefore \, \, \, 2x^2+y^2=\frac{5}{2},corresponding\, foci\, are\, (0, $\pm$1).$