Find a point $P$ on a curve given the gradient of the normal at $P$. The point $P$ lies on curve $y=(x-5)^2$. It is given that the gradient of the normal at $P$ is $-\frac 14$
Find the coordinates of $P$.

First find the gradient function of the curve:
$$\frac{dy}{dx}=2(x-5)$$
We want to find the point where the gradient is $-\frac{1}{-\frac14}=4$
So we solve \begin{align}2(x-5)&=4\\
x-5&=2\\
x&=7\end{align}
And therefore the point $P$ has an $x$ coordinate of $7$, and we can find its $y$ coordinate:
\begin{align}y&=(x-5)^2\\
&=(7-5)^2\\
&=2^2\\
&=4\end{align}
So $P$ has coordinates $(7,4)$
