The locus of the point of trisection of all the double ordinates of the parabola $y^2= lx$ > The locus of the point of trisection of all the double ordinates of the parabola $y^2= lx$ is a parabola whose latus rectum is ? I feel that since the double ordinates are trisected the latus rectum too should be trisected. So I thought the answer should be $\dfrac{l}{3}$. However the given answer says the latus rectum should be $\dfrac{l}{9}$. Am I going wrong somewhere ?

> Am I going wrong somewhere ?

It seems that you think that the focuses of the both parabolas have the same $x$-coordinate.

The coordinates of the points both on the parabola $y^2=lx$ and on $x=t$ are $(t,\pm\sqrt{lt})$.

Since the double ordinates on $x=t$ are trisected, we get $$\left(t,\frac{1\cdot\sqrt{lt}+2(-\sqrt{lt})}{1+2}\right),\quad \left(t,\frac{1\cdot(-\sqrt{lt})+2\cdot \sqrt{lt}}{1+2}\right),$$ i.e. $$\left(t,\pm\frac{\sqrt{lt}}{3}\right)$$ which are on the parabola $y^2=\frac{l}{3^\color{red}{2}}x=\frac{l}{9}x$.