Question

If one vertex of a chord of parabola $y^2=8ax$ is at $(0,0),$ then the locus of a point which divides the chord in the ratio $1:2$ is

• A. $y^2=4ax$
• B. $y^2=8ax$
• C. $y^2=\frac{8a}{3}x$
• D. $y^2=2ax$

Answer 1

The correct option is C $y^2=\frac{8a}{3}x$

Let the point be $P\equiv (h,k)$ and other end point of chord on the parabola be $Q\equiv (x,y)$
Then $h=\frac{x+0}{2+1}$
$\Rightarrow x=3h$
and $k=\frac{y+0}{2+1}$
$\Rightarrow y=3k$

Since $(x,y)$ lies on $y^2=8ax,$
$(3k{)}^2=8a\times 3h$
$\Rightarrow k^2=\frac{8a}{3}h$
Hence, locus of the point $P$ is $y^2=\frac{8a}{3}x$ which is a parabola.