Q31 / 90 · JEE Main 2021 (27 Aug Shift 1) · Mathematics

Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2 t, 0) .$ Let $M$ be the mid-point of $A B$ and the perpendicular bisector of $A B$ meets the $y -$ axis at $C .$ The locus of the mid-point $P$ of $MC$ is

Solution

Mid point $M = (t, 3)$

$m_{AB} = \frac{- 3}{t}$

Perpendicular bisector of $AB$ is

$y - 3 = \frac{t}{3} (x - t)$

So, $C = (0, 3 - \frac{t^{2}}{3})$

Let $P$ be $(h, k)$

$h = \frac{t}{2}; k = 3 - \frac{t^{2}}{6}$

$\Rightarrow k = 3 - \frac{4 h^{2}}{6}$

$\Rightarrow 2 x^{2} + 3 y - 9 = 0$

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