Q31 / 92 · NTA Abhyas (JEE Main) 2020 · Mathematics · Parabola

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

A
$9 x^{2} = y$
B
$3 x^{2} = 2 y$
Correct answer
C
$9 x^{2} = 4 y$
D
$9 x^{2} = 2 y$

Solution

Let $A \equiv (2 t, t^{2}), B \equiv (- 2 t, t^{2})$ be the extremities on the double ordinate $A B .$
Let $C (h, k)$ be it’s trisection point, then
$6 h = 4 t$ 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 $9 x^{2} = 4 y$

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

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Solution

More Parabola Questions

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