Ask Uncle Colin: Messy Geometry

Dear Uncle Colin,

I’m told that the normal to the curve $y=a{x}^2+bx+1$ through the point $(3,1)$ is parallel to the line $3y-x+4=0$ and I need to find $a$ and $b$. There’s a lot going on there and I’m a bit lost!

Bit Unhappy Solving, Yuck!

Hi, BUSY, and thank for your message! There is a lot going on there.

There are two key pieces of information to pull out there:

• The curve passes through $(3,1)$
• The normal to the curve at that point is parallel to a given line.

These are going to give us a pair of equations to solve.

Let’s start with the first piece of information: when $x=3$, $y=1$, so $1=9a+3b+1$. Therefore, $9a+3b=0$ - or $b=-3a$ after a bit of shuffling.

The second piece is more involved. I would start by finding the gradient of the line: rewriting it as $y=\frac{1}{3}x+\frac{4}{3}$ tells you that the gradient is $\frac{1}{3}$.

This is perpendicular to the gradient of the curve, so the curve must have a gradient of $-3$ when $x=3$.

What is the gradient of the curve there? Let’s differentiate: it’s $\frac{dy}{dx}=2ax+b$, so when $x=3$, $-3=6a+b$.

We know that $b=-3a$, so $-3=3a$, which gives $a=-1$ and $b=3$.

A quick check: if $y=-x^2+3x+1$, it does indeed go through $(3,1)$; its derivative is $\frac{dy}{dx}=-2x+3$, which is indeed perpendicular to the line.

Hope that helps!

- Uncle Colin