Question

An ellipse intersects the hyperbola $2x^2-2y^2=1$ orthogonally. The eccentricity of the ellipse is the reciprocal of that of hyperbola. If the axis of the ellipse are along the coordinate axis, then

• A. equation of ellipse is $x^2+2y^2=4$
• B. coordinates of foci of ellipse are $(\pm 1,0)$
• C. equation of ellipse is $x^2+2y^2=2$
• D. coordinates of foci of ellipse are $(\pm \sqrt{2},0)$

Answer 1

Answer: (B), (C)

The correct options are
B coordinates of foci of ellipse are $(\pm 1,0)$
C equation of ellipse is $x^2+2y^2=2$

Let equation of the ellipse with eccentricity $e^{\prime }$ be
$\frac{x^2}{a^{\prime 2}}+\frac{y^2}{b^{\prime 2}}=1$

Equation of the the hyperbola is
$2x^2-2y^2=1\Rightarrow x^2-y^2=\frac{1}{2}$
Hence given hyperbola is an rectangular hyperbola
$e=\sqrt{2},a=\frac{1}{\sqrt{2}}$
$\therefore e^{\prime }=\frac{1}{\sqrt{2}}$
We know that if ellipse and hyperbola are intersecting each other orthogonally then thay are confocal .

So, Coordinates of foci of the ellipse are
$(\pm a^{\prime }{e}^{\prime },0)\equiv (\pm 1,0)\Rightarrow a^{\prime }=\sqrt{2}$
and $b^{\prime 2}=a^{\prime 2}(1-(e^{\prime }{)}^2)=1$

Hence, equation of the ellipse is
$\frac{x^2}{2}+\frac{y^2}{1}=1\Rightarrow x^2+2y^2=2$