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)

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