The Locus of the Midpoint of the Line Segment Joining the Focus to a Moving Point on the Parabola y² = 4ax Is Another Parabola With the Directrix

A standard parabola locus problem, not tied to a specific exam paper.

Problem

The locus of the midpoint of the line segment joining the focus to a moving point on the parabola $y^2=4ax$ is another parabola with the directrix?

Key insight. Writing the moving point in the standard parametric form $(at^2,2at)$ and averaging its coordinates with the fixed focus $(a,0)$ gives $h$ and $k$ purely in terms of the single parameter $t$ — eliminating $t$ between these two expressions produces the equation of the new locus directly.

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Approach

Write a general moving point on the parabola in parametric form $(at^2,2at)$, and the focus as $(a,0)$. Find the midpoint $(h,k)$ of the segment joining these two points, both in terms of $t$. Eliminate $t$ between the two expressions to get the locus equation in $h,k$ (i.e. $x,y$), then identify it as a parabola and read off its directrix.

Solution

Step 1 — Set up the moving point and the focus

Moving point on $y^2=4ax$: $(at^2,2at)$. Focus: $(a,0)$.

Step 2 — Find the midpoint (h, k)

$$h = \frac{a+at^2}{2}, \qquad k = \frac{0+2at}{2} = at$$

Step 3 — Eliminate t

From $k=at$: $t=\dfrac{k}{a}$. Substituting into the expression for $h$:

$$h = \frac{a\left(1+\dfrac{k^2}{a^2}\right)}{2} = \frac{a}{2}+\frac{k^2}{2a}$$

Step 4 — Rearrange into a locus equation

$$2h = a+\frac{k^2}{a} \implies 2h-a = \frac{k^2}{a} \implies k^2 = a(2h-a) = 2ah-a^2$$

$$k^2 = 2a\left(h-\frac{a}{2}\right)$$

Step 5 — Write in x, y and identify the parabola

$$y^2 = 2a\left(x-\frac{a}{2}\right)$$

Comparing with the standard form $y^2=4A(x-x_0)$: $4A=2a \implies A=\dfrac{a}{2}$, with vertex at $x_0=\dfrac{a}{2}$.

Step 6 — Find the directrix

For a parabola $y^2=4A(x-x_0)$, the directrix is at $x=x_0-A$:

$$x = \frac{a}{2}-\frac{a}{2} = 0$$

Answer

Common mistakes

• Forgetting to shift the vertex when identifying $A$ and the directrix. The new parabola isn’t centred at the origin — its vertex sits at $x=\dfrac{a}{2}$, so the directrix formula $x=x_0-A$ (using the shifted vertex $x_0$) must be used, not the simpler $x=-A$ that would apply only if the vertex were at the origin.
• Mixing up which point is the “moving point” and which is the fixed focus when setting up the midpoint. Since the focus $(a,0)$ never changes while the other point sweeps the whole parabola, only the moving point’s coordinates should carry the parameter $t$ — treating both as parametric would give a completely different, incorrect locus.

Practise next

• Find the locus of the midpoint of the segment joining the focus of $y^2=8x$ to a moving point on the parabola, and identify its directrix, using the same parametric elimination approach.