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)

$\frac{1}{4}$

We have: $y=x^2-\frac{1}{x^2}=x^2-x^{-2}$

Now differentiate,

$\frac{dy}{dx}=2x+2x^{-3}$

So slope of tangent at $(-1,0)$ is

$\frac{dy}{dx}{|}_{x=-1}=2(-1)+2(-1{)}^3=-2-2=-4$

Hence slope of normal at the same point is $=\frac{1}{4}$

Note: Normal and tangent are perpendicular to each other