Find a point $P$ on a curve given the gradient of the normal at $P$.

Question

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 $P$.

Answer 1

If the gradient of the normal at P is $\displaystyle -\frac{1}{4}$, the gradient of the tangent to P is the perpendicular, or $4$.

We know the derivative of $(x-5)^2$ to be $2(x-5)$.

If $2(x-5)=4$, $x=7$.

The coordinates are then $(7,4)$.

Have a nice day, finish your homework :D

Answer 2

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 $(7,4)$