I've been asked to prove that the normal to the curve $y=2x^2 - 3x^{-1/2}$ at the point $(1,-1)$ passes through the point $(12,3)$.
$\frac{dy}{dx} = 4x + \frac{3}{2}x^{-3/2}$
Hence, at the point $(1,-1)$, the gradient of the tangent $=\frac{11}{2}$
Therefore, the gradient of the normal at that point$=-\frac{2}{11}$
So, the equation for the normal is $$y+1=-\frac{2}{11}(x-1)$$
$$11y=-2x-9$$
However, when I substitute the value $12$ in for $x$ I get $y=-3$, rather than $3$, so that equation can't be right.
What have I done wrong?