The slope of the normal to the curve $y=2 x^2-4$ at $P(1,-2)$ is :

$-\frac{1}{4}$

$y=2 x^2-4$ at $(P(1,-2))$

slope of tangent at $P=\frac{d y}{d x}=\frac{d}{d x}\left(2 x^2-4\right)$

$\Rightarrow \frac{d y}{d x} =4 x$

so $\left.\frac{d y}{d x}\right]_{(1,-2)}=4(1)=4$

So we know that at any point 

slope of tangent (m) × slope of normal (n) = -1 

so $4 \times n=-1$

$n=-\frac{1}{4}$  (slope of normal at P)