# conic sections – Tangent and normal to a curve at a minimum

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?