Question

If $4(x-\sqrt{2}{)}^2+\lambda (y-\sqrt{3}{)}^2=45$ and $(x-\sqrt{2}{)}^2-4(y-\sqrt{3}{)}^2=5$ cut orthogonally, then $\lambda$ is equal to ______.

Answer 1

Answer: 9

Here, ellipse $\frac{(x-\sqrt{2}{)}^2}{45/4}+\frac{(y-\sqrt{3}{)}^2}{45/\lambda }=1$
and hyperbola $\frac{(x-\sqrt{2}{)}^2}{5}-\frac{(y-\sqrt{3}{)}^2}{5/4}=1$ cut orthogonally.
So, conics will be confocal.
$\therefore \frac{45}{4}-\frac{45}{\lambda }=5+\frac{5}{4}$
$\therefore \lambda =9$