Question

The slope of the normal to the curve $y=x^2-\frac{1}{x^2}$ at $\left(-1,0\right)$ is

• A. $\frac{1}{4}$
• B. $-\frac{1}{4}$
• C. 4
• D. -4
• E. 0

Answer 1

Answer: (A)

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
$=-\frac{1}{dy/dx}$
$=-\frac{1}{-4}=\frac{1}{4}$