Review question

# Can we find the locus of a midpoint created by a parabola? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8399

## Solution

Show that the tangent to the curve $y = (x+k)^2$ at the point where $x = 2k$ is $y + 3k^2 = 6kx.$

We have $y = (x+k)^2 = x^2+2kx+k^2$, and so $\dfrac{dy}{dx}= y' = 2x + 2k$. (If you know the chain rule, you can alternatively use this here.)

This tells us that when $x = 2k, y = 9k^2$, and $y' = 6k$. So the gradient of the tangent is $6k$.

We can write the equation of the tangent as $y = 6kx+c$, and we know that the point $(2k, 9k^2)$ is on the line, so $9k^2 = 12k^2 +c \implies c = -3k^2 .$

Thus the equation is $y = 6kx - 3k^2$ or, after rearranging, $$$\label{eq:given-form-for-tangent} y + 3k^2 = 6kx,$$$

as required.

This tangent meets the $x$-axis at $P$ and the $y$-axis at $Q$. The mid-point of $PQ$ is $M$.

Find the co-ordinates of $M$ in terms of $k$ and hence deduce the equation of the locus of $M$ as $k$ varies.

From $\eqref{eq:given-form-for-tangent}$, the coordinates of $Q$ are just $(0,-3k^2)$. Similarly, the $x$-coordinate of $P$ is given by $0 = 6kx - 3k^2,$ and so $x = \dfrac{k}{2}$. Thus, the coordinates of $P$ are $\left( \dfrac{k}{2}, 0 \right)$.

The midpoint of the line segment $PQ$ is thus $\left(\dfrac{0+k/2}{2}, \dfrac{-3k^2+0}{2} \right) = \left( \dfrac{k}{4}, -\dfrac{3k^2}{2} \right).$

To find the locus of $M$, we need to eliminate $k$ here. So if $(x,y)$ is on the locus of $M$, $x = \dfrac{k}{4} \implies k = 4x,\quad y = -\dfrac{3k^2}{2} \implies y = -24x^2.$

Now we have the locus of $M$ as $k$ varies; a vertex-up parabola, with its maximum at the origin. You can observe the locus in the following applet by varying $k$.