Question

The slope of the 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. $-4$
• E. $0$

Answer 1

Answer: (A)

$\frac{1}{4}$

Given curve is $y=x^2-\frac{1}{x^2}$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=2x+\frac{2}{x^3}$
At point $(-1,0)$.
$\frac{dy}{dx}=2(-1)+\frac{2}{(-1{)}^3}=-4$
Therefore, slope of normal to the curve
$=-\frac{1}{dy/dx}$
$=-\frac{1}{-4}=\frac{1}{4}$