When does the normal to a parabola pass through this point?

When does the normal to a parabola pass through this point?

Question

As shown in the diagram below: $C$ is the parabola with equation $y=x^2$; $P$ is the point $(0,1)$; $Q$ is the point $(a,a^2)$ on $C$; $L$ is the normal to $C$ which passes through $Q$.

1. Find the equation of $L$.

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1. For what values of $a$ does $L$ pass through $P$?

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1. Determine $\big |QP\big |^2$ as a function of $a$, where $\big |QP\big |$ denotes the distance from $P$ to $Q$.

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1. Find a point $R$, in the $xy$-plane but not on $C$, such that $\big |RQ\big |$ is smallest for a unique value of $a$. Briefly justify your answer.

Interactive graph

As shown in the diagram below: $C$ is the parabola with equation $y=x^2$; $P$ is the point $(0,1)$; $Q$ is the point $(a,a^2)$ on $C$; $L$ is the normal to $C$ which passes through $Q$.

1. For what values of $a$ does $L$ pass through $P$?

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Solution

Firstly we need to find the gradient of $C$ at $Q$. To do this we differentiate $y=x^2$ to get $\frac{dy}{dx}=2x,$ and so at $Q$ the gradient of $C$ is $2a$.

Now we use the fact that (gradient of normal) $\times$ (gradient of tangent) $= -1$, so the gradient of $L$ is $-\dfrac{1}{2a}$.

BUT beware! This causes trouble when $a=0$, so we’ll have to consider this case separately.

Now $L$ is a straight line of the form $y=mx+c$, where $m$ is the gradient and $c$ is a constant.

So we substitute in $y=a^2$, $x=a$ and $m=-\dfrac{1}{2a}$ to find that $c=a^2 + \dfrac{1}{2}$, and the equation of $L$ is $y=-\frac{1}{2a} x + \frac{1}{2} + a^2, \quad\quad (a \neq 0),$ and when $a=0$, the coordinates of $Q$ are $(0,0)$ and the gradient at $Q$ is $0$, so the equation of the normal $L$ is $x=0.$

First we should notice that $a=0$ is a solution, as the line $x=0$ passes through $P$.

Then we substitute the coordinates of $P$ into the equation for $L$ ($a \neq 0$) to find the other possible values for $a$.

So $1=-\frac{1}{2a}(0) + \frac{1}{2} + a^2,$ which we can rearrange to get $a^2=\frac{1}{2},$ and so our solutions are $a=\pm \frac{1}{\sqrt{2}}, 0.$

We can use Pythagoras here, giving

$$\big |QP\big |^2 = (a^2-1)^2 + (a-0)^2 = a^4-2a^2 +1 +a^2 = a^4 - a^2 +1.$$

Note that $\dfrac{d}{da}\big |QP\big |^2 = 4a^3-2a = 2a(2a^2-1) = 0$ when $a=\pm \dfrac{1}{\sqrt{2}}, 0$.

Thus the points on the parabola we’ve identified where $L$ goes through $P$ are stationary points for $\big |QP\big |^2$, and thus for $\big |QP\big |$.

There are lots of possible answers!

Perhaps the simplest example is a point of the form $(0,-b)$ for some positive $b$, since it’s easy to see that the unique point on $C$ which it is closest to is the origin.

Any point $R$ on $L$ and ‘outside’ the parabola will count, since $Q$ will be the unique point closest to $R$ (because the parabola’s gradient is always increasing).

But as $Q$ moves around the parabola, any point ‘outside’ $C$ will be on some normal through $Q$, and so will be a possible $R$.

In fact, the only points not closest to a unique point on $C$ (and which are not on $C$ itself) are the points on the positive $y$-axis.