The locus of the mid-point of the line segment joining the focus to a moving point on the parabola, $y^2=4 a x$ is a conic. The equation of the directrix of that conic is

• A

  $y=a$
• B

  $x=a$
• C

  $y=0$
• Correct answer
  D

  $x=0$

Solution

Let $Q(h, k)$ be the mid-point of line joining the focus $F(a, b)$ and variable point $p\left(x_0, y_0\right)$.
$$ \begin{array}{ll} \therefore (h, k)=\left(\frac{x_0+a}{2}, \frac{y_0}{2}\right) \\ \Rightarrow h=\frac{x_0+a}{2} \text { and } k=\frac{y_0}{2} \\ \Rightarrow & x_0=2 h-a \text { and } y_0=2 k \\ \text { Since, } P\left(x_0, y_0\right) \text { lies on parabola } y^2=4 a x \\ \therefore & y_0^2=4 a x_0 \Rightarrow(2 k)^2=4 a(2 h-a) \\ \Rightarrow & 4 k^2=4 a(2 h-a) \Rightarrow k^2=2 a\left(h-\frac{a}{2}\right) \end{array} $$ Since, $P\left(x_0, y_0\right)$ lies on parabola $y^2=4 a x$ $$ \begin{array}{lrl} \therefore & y_0^2=4 a x_0 & \Rightarrow(2 k)^2=4 a(2 h-a) \\ \Rightarrow & 4 k^2=4 a(2 h-a) & \Rightarrow k^2=2 a\left(h-\frac{a}{2}\right) \end{array} $$ which is equation of parabola, $$ \therefore \quad y^2=2 a\left(x-\frac{a}{2}\right) $$ $\therefore$ Equation of directrix is given by $$ x-\frac{a}{2}=\frac{-a}{2} \Rightarrow x=0 $$

Hence, the correct answer is option: 'D'.