Question

The intersection angle of the curve $xy=a^2$ and $x^2-y^2=a^2$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{2}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{2}$

$xy=a^2$ and $x^2-y^2=a^2$

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

$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$ and $\frac{dy}{dx}=\frac{x}{y}$
$\therefore {\left(\frac{dy}{dx}\right)}_1=-\frac{y}{x}$ and ${\left(\frac{dy}{dx}\right)}_2=\frac{x}{y}$

	$\Rightarrow {\left(\frac{dy}{dx}\right)}_1\times {\left(\frac{dy}{dx}\right)}_2=-\frac{y}{x}\times \frac{x}{y}=-1$

Hence, the intersection angle $=\frac{\pi }{2}$.