Question

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

• A. $3x^2+2y-6=0$
• B. $2x^2+3y-9=0$
• C. $3x^2-2y-6=0$
• D. $2x^2-3y+9=0$

Answer 1

The correct option is B $2x^2+3y-9=0$

Let $A(0,6)$ and $B(2t,0)$
Then, mid point of $AB$ is $M=(t,3)$
and slope of $AB$ is
$m_{AB}=\frac{-6}{2t}=-\frac{3}{t}$
$\therefore$ Equation of perpendicular bisector of $AB$ is
$y-3=\frac{t}{3}(x-t)$
$\Rightarrow 3y-9=tx-t^2$
Since perpendicular bisector of $AB$ cuts the $y-$axis.
$\therefore C=(0,\frac{9-t^2}{3})$
Let mid point of $MC$ be $(h,k)$
Then, $\frac{t+0}{2}=h$
$\Rightarrow t=2h$
and $\frac{3+\frac{9-t^2}{3}}{2}=k$
$\Rightarrow 9-t^2=6k-9$
$\Rightarrow 4h^2+6k-18=0(\because t=2h)$
$\therefore$ locus of $(h,k)$ is $4x^2+6y-18=0$
$\therefore 2x^2+3y-9=0$