Question

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

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

Answer 1

The correct option is C $x=0$

Let the focus of the parabola $y^2=4ax$ is $S:(a,0)$

Let $P:(a{t}^2,2at)$ be any point on the parabola, then

Coordinates of the mid point of $SP$ are:

$x=\frac{a(t^2+1)}{2},y=\frac{2at+0}{2}$

Eliminate $t$ to get locus of the mid point

$y^2=2ax-a^2\Rightarrow y^2=2a(x-\frac{a}{2}).........\left(1\right)$

which is a parabola of the form $Y^2=4AX............\left(2\right)$

$Y=y,$ $X=x-\frac{a}{2},A=\frac{a}{2}$

Equation of the directrix of $\left(2\right)$ is $X=-A$

So, equation of the directrix of $\left(1\right)$ is $x-\frac{a}{2}=-\frac{a}{2}\Rightarrow x=0$