Let A be a fixed point

Question:

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

  1. $3 x^{2}-2 y-6=0$

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

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

  4. $2 x^{2}-3 y+9=0$


Correct Option: , 3

Solution:

$\mathrm{A}(0,6)$ and $\mathrm{B}(2 \mathrm{t}, 0)$

Perpendicular bisector of $\mathrm{AB}$ is

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

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

Let P be (h,k)

$\mathrm{h}=\frac{\mathrm{t}}{2} ; \mathrm{k}=\left(3-\frac{\mathrm{t}^{2}}{6}\right)$

$\Rightarrow \mathrm{k}=3-\frac{4 \mathrm{~h}^{2}}{6} \Rightarrow 2 \mathrm{x}^{2}+3 \mathrm{y}-9=0$ option (3)

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