Question

The angle of intersection of the two curves $xy = a^2$ and $x^2 + y^2 = 2a^2$ is

• A. $\frac {\pi} {6}$
• B. $\frac {\pi} {4}$
• C. $\frac {\pi} {3}$
• D. none of these

Answer 1

The Correct Option is D

$xy = a^2$ $\Rightarrow y = \frac{a^{2}}{x}$ $\Rightarrow \frac{dy}{dx}=-\frac{a^{2}}{x^{2}}$ $x^{2}+y^{2} = 2a^{2}$ $\Rightarrow 2x + 2y \frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$. Let $\left(x_{1}, y_{1}\right)$ be the pt. of intersection of the curves. $\left(\frac{dy}{dx}\right)$ at $\left(x_{1}, y_{1}\right)$ for first curve $= - \frac{a^{2}}{x^{2}_{1}}$ and for second $= - \frac{x_{1}}{y_{1}}$. Now $x^{2}_{1} + y^{2}_{1} = 2a^{2}, x_{1}y_{1} = a^{2}$ $\Rightarrow \left(x^{2}_{1}-y^{2}_{1}\right)^{2}=\left(x^{2}_{1}-y^{2}_{1}\right)^{2}-4\,x^{2}_{1}y^{2}_{1}$ $= 4a^{4} - 4a^{4} = 0$ $\Rightarrow x^{2}_{1}=y^{2}_{1}$ But $x^{2}_{1}+y^{2}_{1}=2a^{2}$ $\Rightarrow 2\,x^{2}_{1} = 2a^{2}$ $\Rightarrow x^{2}_{1} = a^{2}$ $\Rightarrow y^{2}_{1} = a^{2}$ $\therefore x_{1} = y_{1} = a$ $\therefore$ slopes of the two curves at $\left(x_{1}, y_{1}\right)$ are equal and is $- 1$. $\therefore$ angle $= \pi$