Review question

# A point $Y$ moves to define a point $P$ — can we find the locus of $P\,$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6583

## Solution

$X$ is a fixed point $(3,0)$, $Y$ is the point $(0,a)$. A line is drawn through $Y$ and perpendicular to $XY$, meeting the $x$-axis at $Q$. On this line a point $P$ is taken so that $Y$ is the mid-point of $QP$. Find the coordinates of $P$.

We begin by sketching a graph of the situation.

Our strategy is now clear: we will find the gradient of the line $XY$, and so be able to work out the gradient of $PQ$. This will allow us to find the equation of $PQ$, and hence the coordinates of $Q$. Since $Y$ is the midpoint of $PQ$, we will then be able to find the coordinates of $P$.

The gradient of the line segment $XY$ is $\frac{a-0}{0-3} = -\frac{a}{3}.$ Provided that $a \ne 0$, the equation of any line perpendicular to $XY$ must therefore have a gradient equal to $\dfrac{3}{a}$. (We will deal with $a=0$ later on as a special case.)

So the equation of the line with gradient $\dfrac{3}{a}$ that passes through $Y(0,a)$ is $y = \dfrac{3}{a} x + a,$ since the $y$-intercept is $a$.

The point where this line intersects the $x$-axis ($y=0$) has $x$-coordinate $-\dfrac{a^2}{3}$, so the point $Q$ has coordinates $\left( -\dfrac{a^2}{3}, 0 \right)$.

To get from $Q$ to $Y$ we go $\dfrac{a^2}{3}$ to the right, then $a$ up, so we need to repeat this to get to $P$, which is therefore the point $\left(\dfrac{a^2}{3}, 2a \right)$.

It is also clear from the sketch that the $x$-coordinates of $P$ and $Q$ are negatives of each other, and that the $y$-coordinate of $Q$ is twice the $y$-coordinate of $Y$.

In the case that $a = 0$, the points $Y, Q$ and $P$ coincide at the origin. So for all values of $a$, the point $P$ can be expressed as $\left( \dfrac{a^2}{3}, 2a \right)$.

Is this still true if $a<0$?

Find the equation of the locus of $P$ for varying values of $a$.

We need to determine the locus of the point $P\left( \dfrac{a^2}{3}, 2a \right)$ as $a$ varies.

If $x = \dfrac{a^2}{3}$ and $y = 2a$, then $a = y/2$, giving $x = \dfrac{(y/2)^2}{3}$. Rearranging yields $y^2 = 12x$, which is the equation of a parabola which is symmetrical about the $x$-axis.

This GeoGebra applet can help us to see what is going on.