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

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$3 x^{2} + 2 y - 6 = 0$

$3 x^{2} - 2 y - 6 = 0$

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

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

answer is C.

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Detailed Solution

Given, A(0, 6) and B(2t, 0)

Mid-point of AB = M(t, 3)

$s l o p e o f A B = \frac{6 - 0}{0 - 2 t} = - \frac{3}{t} \Rightarrow s l o p e o f p e r p e n d i c u l a r b i s e c t o r = \frac{t}{3}$

Equation of perpendicular bisector of AB passes through M.

$\begin{array}{l} ∴ y - 3 = \frac{t}{3} (x - t) . . . (i) \\ So, C (0, 3 - \frac{t^{2}}{3}) \end{array}$

Intersection of Eq. (i) on Y-axis

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

Let mid-point of MC is (h, k).

$\begin{array}{l} Then, (h, k) = (\frac{t}{2}, 3 - \frac{t^{2}}{6}) \\ \Rightarrow h = \frac{t}{2}, k = 3 - \frac{t^{2}}{6} \\ Eliminating t, we get \\ 2 h^{2} = 3 (3 - k) \\ Locus of (h, k) \\ 2 x^{2} = 3 (3 - y) \\ \Rightarrow 2 x^{2} + 3 y - 9 = 0 \end{array}$