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
$\frac{d y}{d x}=2 x+\frac{2}{x^{3}}$
At point $(-1,0)$
$\frac{d y}{d x}=2(-1)+\frac{2}{(-1)^{3}}=-4$
$\therefore $ Slope of normal to the curve
$=-\frac{1}{d y / d x} $
$=-\frac{1}{-4}=\frac{1}{4}$