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 $=\sqrt{2}$
$\Rightarrow \frac{x^2}{\left(\frac{1}{2}\right)} - \frac{y^2}{\left(\frac{1}{2}\right)} = 1 ....$ (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}=2 b^{2} $
$\Rightarrow x^{2}+2 y^{2}=2 b^{2} ....$(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
$4 x-4 y \frac{d y}{d x} =0 \Rightarrow \frac{d y}{d x}=\frac{x}{y} $
$\left.\frac{d y}{d x}\right|_{\text {at } A} =\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta$
and on differentiating Eq. (ii), we get
$2 x+4 y \frac{d y}{d x}=\left.0 \Rightarrow \frac{d y}{d x}\right|_{\text {at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} \theta$
Since, ellipse and hyperbola are orthogonal.
$\therefore -\frac{1}{2} \operatorname{cosec}^{2} \theta=-1 $
$\Rightarrow \operatorname{cosec}^{2} \theta=2 \Rightarrow \theta=\pm \frac{\pi}{4}$
$\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}=2 b^{2}$
$\Rightarrow b^{2}=1$
Equation of ellipse is $x^{2}+2 y^{2}=2$.
Coordinates of foci $(\pm a e, 0)=\left(\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0\right)=(\pm 1,0)$
If major axis is along $Y$-axis, then
$\frac{1}{\sqrt{2}} =\sqrt{1-\frac{a^{2}}{b^{2}}} \Rightarrow b^{2}=2 a^{2} $
$\therefore 2 x^{2}+y^{2} =2 a^{2} \Rightarrow Y^{\prime}=-\frac{2 x}{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 \operatorname{cosec} \theta=-1 $
$\Rightarrow \operatorname{cosec}^{2} \theta=\frac{1}{2} \Rightarrow \theta=\pm \frac{\pi}{4} $
$\therefore 2 x^{2}+y^{2}=2 a^{2} $
$\Rightarrow 2+\frac{1}{2}=2 a^{2} \Rightarrow a^{2}=\frac{5}{4} $
$\therefore 2 x^{2}+y^{2}=\frac{5}{2}, $ corresponding foci are $(0, \pm 1)$ .