Question

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

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

Answer 1

The correct option is A $\frac{1}{4}$

Slope of the tangent of the curve at $(-1,0)$ is
$\frac{dy}{dx}{|}_{x=-1}=2x+\frac{2}{x^3}{|}_{x=-1}=-4\therefore \text{slope of normal to the curve}=-\frac{1}{\frac{dy}{dx}}=\frac{1}{4}$