JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Ellipse

The hyperbola is $\frac{x^2}{1/2}-\frac{y^2}{1/2}=1$, so $a^2=b^2=1/2$ and eccentricity $e_h=\sqrt{1+1}=\sqrt{2}$.

The ellipse has eccentricity $e_e=\frac{1}{\sqrt{2}}$.

For orthogonal intersection, the ellipse and hyperbola are confocal. The foci of the hyperbola are at $(\pm 1,0)$.

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$, we have $ae=1$ where $e=\frac{1}{\sqrt{2}}$.

Thus $a=\sqrt{2}$ and $b^2=a^2(1-e^2)=2(1-1/2)=1$.

The equation is $\frac{x^2}{2}+\frac{y^2}{1}=1$ or $x^2+2y^2=2$.

The foci are at $(\pm ae,0)=(\pm 1,0)$.