[Math] Prove that the normal to a quadratic curve passes through a specific point

Question

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?

Answer 1

You are correct, the question is wrong (or printed incorrectly). Look at this graph:

The red line is the graph of the equation in the question, the green line is the graph of the equation of the normal. It is clear that your normal line is correct for this graph.