Question

The locus of the trisection point of any arbitrary double ordinate of the parabola $x^{2}=4y,$ is

• A. $9x^{2}=y$
• B. $3x^{2}=2y$
• C. $9x^{2}=4y$
• D. $9x^{2}=2y$

Answer 1

Answer: (C)

Let $A\equiv \left(2 t , t^{2}\right),B\equiv \left(- 2 t , t^{2}\right)$ be the extremities on the double ordinate $AB.$
Let $C\left(h , k\right)$ be it’s trisection point, then
$6h=4t$ and $k=t^{2}$
$\Rightarrow t=\frac{3 h}{2},t^{2}=k\Rightarrow k=\frac{9 h^{2}}{4}$
Thus, locus of $C$ is $9x^{2}=4y$