IIT-JEE 2009 · Paper 2, Mathematics question 25

Hyperbola. $\dfrac{x^2}{1/2}-\dfrac{y^2}{1/2}=1$ is rectangular: $e=\sqrt2$, foci $(\pm1,0)$.

Ellipse. $e=\frac1{\sqrt2}$, so $b^2=\frac{a^2}2$.

Orthogonality. An ellipse and a hyperbola with the same centre and axes cut at right angles exactly when they are confocal ($a^2-b^2=A^2+B^2$). So $\frac{a^2}2=1$: $a^2=2$, $b^2=1$.

Ellipse: $x^2+2y^2=2$, with foci $(\pm1,0)$.

(A), (B)