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

$x=0$

Explanation for the correct answer:

The focus of the parabola $y^2=4ax$ is $\left(a,0\right)$.

Let $Q\left(x,y\right)$ be any point on the parabola and $P\left(h,k\right)$ be the midpoint of the focus and the point on the parabola.

By the midpoint formula

$\Rightarrow$$\left(h,k\right)=\left(\frac{x+a}{2},\frac{y}{2}\right)$

$\Rightarrow$ $x=2h-a$

$\Rightarrow$ $y=2k$

Substituting these values of $x$ and $y$ in the equation of the parabola we get

$\Rightarrow$${\left(2k\right)}^2=4a\left(2h-a\right)$

$\Rightarrow$ $4k^2=8ah-4a^2$

$\Rightarrow$ $k^2=2ah-a^2$

To find the locus of point $P$ replace $h$ with $x$ and $k$ with $y$

$\Rightarrow$$y^2=2ax-a^2$

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

The directrix of the parabola is given as

$x-\frac{a}{2}=\frac{-2a}{4}$ $...[\because y^2=4ax$ has directrix $x=-a]$

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

$\Rightarrow$ $x=0$

Hence, option $\left(C\right)$ is the correct answer.