Dear Uncle Colin,

I’m told that the normal to the curve $y=ax^2 + bx + 1$ 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 $\dydx = 2ax + b$, 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 $\dydx = -2x + 3$, which is indeed perpendicular to the line.

Hope that helps!

- Uncle Colin