Question

Slope of Normal to the curve $y=x^2 - \frac {1}{x^2}$ at $(-1,0)$ is

• A. $\frac {1}{4}$
• B. $- \frac {1}{4}$
• C. $4$
• D. $-4$

Answer 1

Answer: (A)

$\frac {1}{4}$

Answer 2

We have, $\quad y=x^{2}-\frac{1}{x^{2}}$ $\Rightarrow \frac{dy}{dx}=2x+\frac{2}{x^{3}}$ On putting x = -1, we get $\frac{dy}{dx}=-2-2=-4$ $\therefore$ Slope of normal $= \frac{-1}{\left(\frac{dy}{dx}\right)}=\frac{-1}{-4}=\frac{1}{4}$