The slope of the normal to the curve y = 2x^2 at x = 1 is:

The correct answer is Option (4) → -1/4

Given curve $y=2x^2$

Slope of tangent

$\frac{dy}{dx}=4x$

At $x=1$

$m_{\text{tangent}}=4$

Slope of normal

$m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}$

$=-\frac{1}{4}$

The slope of the normal at $x=1$ is $-\frac{1}{4}$.