The angle between three curves $xy =\sqrt{2}$ and $x^2-y^2=1$ at $(\sqrt{2}, 1)$ is :

90°

The correct answer is Option (4) → 90°

$xy =\sqrt{2}$

$x^2-y^2=1$

$P(\sqrt{2}, 1)$

Differentiating both equation wrt x

$y+x\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{y}{x}⇒\left.\frac{dy}{dx}\right]_{(\sqrt{2}, 1)}=-\frac{1}{\sqrt{2}}$

$2x-2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{x}{y}$

$\left.\frac{dy}{dx}\right]_{(\sqrt{2}, 1)}=\sqrt{2}$

so $-\frac{1}{\sqrt{2}}×\sqrt{2}=-1$

so both are inclined at 90°