Question

The locus of the midpoints of all chords of the parabola $y^2=4ax$ through its vertex is another parabola with directrix

• A. $x=-a$
• B. $x=a$
• C. $x=0$
• D. $x=-\frac{a}{2}$

Answer 1

The correct option is D $x=-\frac{a}{2}$

As seen in the figure alongside, a point on the parabola $y^2=4ax$ can be written as $(a{t}^2,2at)$

Since the chord passes through the vertex, which is $(0,0)$, the midpoint is given by $(\frac{a{t}^2}{2},at)$

Let $h=\frac{a{t}^2}{2}$ and $k=at$

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

$\Rightarrow 2ah=k^2$

Replacing $h$ by $x$ and $k$ by $y$, we get the required locus as $y^2=2ax$ i.e. $y^2=4\left(\frac{a}{2}\right)x$

Thus, the directrix is given by $x=-\frac{a}{2}$