Question

An ellipse intersects the hyperbola $2x^2-2y^2=1$ 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 $x^2+2y^2=2$
• B. the foci of the ellipse are $(\pm 1,0)$
• C. the equation of the ellipse is $x^2+2y^2=4$
• 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
$\frac{x^2}{2}+\frac{y^2}{1}=1$