Question

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

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

Answer 1

Answer: (A), (B)

The ellipse and hyperbola will be confocal.
Foci of given hyperbola are $(\pm 1,0)$ :
Let equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
$\therefore\left(\pm a \times \frac{1}{\sqrt{2}}, 0\right) \equiv(\pm 1,0)$
or $a=\sqrt{2}$ and $e=\frac{1}{\sqrt{2}}$
or $b^{2}=a^{2}\left(1-e^{2}\right)$
or $b^{2}=1$
Therefore, the equation of the ellipse is
$\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$