Find the points P on the curve with equation $y=x^2-2$ such that the normal to the curve at P passes through the point (0,0).

I said

$y=x^2-2$

$frac{dy}{dx} = 2x rightarrow$ gradient of normal = $frac{-1}{2x}$

Equation of normal: $y-0=frac{-1}{2x}(x-0)rightarrow y = frac{-1}{2}$

At P, $x^2-2=frac{-1}{2}rightarrow$ coordinates are $pmsqrtfrac{3}{2},-frac{1}{2}$

But the function $y=x^2-2$ has a minimum at $0,-2$ and I know the gradient here is 0 and the the normal to this passes through (0,0)but my calculation method does not pick this up.

Also, the gradient of the curve at (0,0) is 0 and that of the normal is $infty$ so these cannot be multiplied together to get -1. Is the “rule” inappropriate here?