# Prove that the normal to a quadratic curve passes through a specific point

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?

• Looking at that graph using a graphing calculator leads me to believe that you are correct and the question is wrong. May 13, 2015 at 20:59
• I am lead to that conclusion @SufyanNaeem, although I can't see how it is wrong May 13, 2015 at 21:02
• Thanks @imulsion - I thought that might be the case. Certainly, I can't find a flaw in my logic. May 13, 2015 at 21:02
• @SufyanNaeem - I take it that you mean therefore that the initial quadratic is wrong? May 13, 2015 at 21:06 