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

Answer: (A), (B)

Eccentricity of hyperbola $=\sqrt{2}$
$\Rightarrow \frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=\mathrm{1....}$ (i)
Eccentricity of hyperbola $=\sqrt{2}$
So, eccentricity of ellipse $=1/\sqrt{2}$
Let equation of ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[$ 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
$A\left(\frac{1}{\sqrt{2}}\sec \theta ,\frac{1}{\sqrt{2}}\tan \theta \right)$
On differentiating Eq. (i), we get
$4x-4y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x}{y}$
${\frac{dy}{dx}|}_{\text{at }A}=\frac{\sec \theta }{\tan \theta }=\mathrm{cosec}\theta$
and on differentiating Eq. (ii), we get
$2x+4y\frac{dy}{dx}={0\Rightarrow \frac{dy}{dx}|}_{\text{at }A}=-\frac{x}{2y}=-\frac{1}{2}\mathrm{cosec}\theta$
Since, ellipse and hyperbola are orthogonal.
$\therefore -\frac{1}{2}{\mathrm{cosec}}^2\theta =-1$
$\Rightarrow {\mathrm{cosec}}^2\theta =2\Rightarrow \theta =\pm \frac{\pi }{4}$
$\therefore A\left(1,\frac{1}{\sqrt{2}}\right)$ or $\left(1,-\frac{1}{\sqrt{2}}\right)$
Form Eq. (i), $1+2{\left(\frac{1}{\sqrt{2}}\right)}^2=2b^2$
$\Rightarrow b^2=1$
Equation of ellipse is $x^2+2y^2=2$.
Coordinates of foci $(\pm ae,0)=\left(\pm \sqrt{2}\cdot \frac{1}{\sqrt{2}},0\right)=(\pm 1,0)$
If major axis is along $Y$-axis, then
$\frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}}\Rightarrow b^2=2a^2$
$\therefore 2x^2+y^2=2a^2\Rightarrow Y^{\prime }=-\frac{2x}{y}$
$\Rightarrow y^{\prime }\left(\frac{1}{\sqrt{2}}\sec \theta ,\frac{1}{\sqrt{2}}\tan \theta \right)=\frac{-2}{\sin \theta }$
As ellipse and hyperbola are orthogonal
$\therefore -\frac{2}{\sin \theta }\cdot \mathrm{cosec}\theta =-1$
$\Rightarrow {\mathrm{cosec}}^2\theta =\frac{1}{2}\Rightarrow \theta =\pm \frac{\pi }{4}$
$\therefore 2x^2+y^2=2a^2$
$\Rightarrow 2+\frac{1}{2}=2a^2\Rightarrow a^2=\frac{5}{4}$
$\therefore 2x^2+y^2=\frac{5}{2},$ corresponding foci are $(0,\pm 1)$ .