An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is…
An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then
equation of ellipse is $x^2+2 y^2=2$
the foci of ellipse are $(\pm 1,0)$
equation of ellipse is $x^2+2 y^2=4$
the foci of ellipse are $(\pm \sqrt{2}, 0)$
Solution
Given, $2 x^2-2 y^2=1$
$
\Rightarrow \quad \frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=1
$
Eccentricity of hyperbola $=\sqrt{2}$
So, eccentricity of ellipse $=1 / \sqrt{2}$
Let equation of ellipse be
$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \\
& \Rightarrow \quad \frac{b^2}{a^2}=\frac{1}{2} \Rightarrow a^2=2 b^2 \\
& \therefore \quad x^2+2 y^2=2 b^2 \\
&
\end{aligned}
$
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),
$
4 x-4 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x}{y}
$
$
\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } A}=\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta
$
On differentiating Eq. (ii),
$
2 x+4 y \frac{d y}{d x}=0
$
$
\therefore \quad\left(\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.
$
\begin{aligned}
& \therefore \quad-\frac{1}{2} \operatorname{cosec}^2 \theta=-1 \\
& \Rightarrow \quad \operatorname{cosec}^2 \theta=2 \Rightarrow \theta=\pm \frac{\pi}{4} \\
& \therefore A\left(1, \frac{1}{\sqrt{2}}\right) \text { or }\left(1,-\frac{1}{\sqrt{2}}\right)
\end{aligned}
$
From Eq. (i), $1+2\left(\frac{1}{\sqrt{2}}\right)^2=2 b^2$ $\Rightarrow \quad b^2=1$
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\right)=(\pm 1,0)
$
Hence, options (a) and (b) are correct. If major axis is along $Y$-axis, then
$
\begin{aligned}
& \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 \quad y^{\prime}\left(\frac{1}{\sqrt{2}} \sec \theta \frac{1}{\sqrt{2}} \tan \theta\right)=\frac{-2}{\sin \theta} \\
&
\end{aligned}
$
As ellipse and hyperbola are orthogonal.
$
\begin{array}{ll}
\therefore & -\frac{2}{\sin \theta} \cdot \operatorname{cosec} \theta=-1 \\
\Rightarrow & \operatorname{cosec}^2 \theta=1 \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} \Rightarrow 2 x^2+y^2=\frac{5}{2}
\end{array}
$
Corresponding foci are $(0, \pm 1)$