Coordinates of the given points

Fixed point: $A(0,6)$

Moving point: $B(2t,0)$

Mid-point $M$ of $AB$

Using midpoint formula, $M\left(\frac{0+2t}{2},\frac{6+0}{2}\right)=(t,3)$

Slope of $AB$

$m_{AB}=\frac{0-6}{2t-0}=\frac{-6}{2t}=\frac{-3}{t}$

Therefore, slope of the perpendicular bisector of $AB$ is $m_\perp=\frac{t}{3}$

Equation of the perpendicular bisector of $AB$

It passes through $M(t,3)$, so $y-3=\frac{t}{3}(x-t)$

Point $C$ where this bisector meets the $y$-axis

On the $y$-axis, $x=0$

Substitute into the line equation: $y-3=\frac{t}{3}(0-t)= -\frac{t^2}{3}$ $y=3-\frac{t^2}{3}$

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

Mid-point $P$ of $MC$

Let $P(x,y)$ be the midpoint of $M(t,3)$ and $C\left(0,3-\frac{t^2}{3}\right)$.

Then, $x=\frac{t+0}{2}=\frac{t}{2}$ $y=\frac{3+\left(3-\frac{t^2}{3}\right)}{2}=\frac{6-\frac{t^2}{3}}{2}=3-\frac{t^2}{6}$

Eliminate the parameter $t$

From $x=\frac{t}{2} \Rightarrow t=2x$

Substitute into the expression for $y$: $y=3-\frac{(2x)^2}{6}=3-\frac{4x^2}{6}=3-\frac{2x^2}{3}$

Multiply by 3: $3y=9-2x^2$

Rearranging, $2x^2+3y-9=0$

Match with the options

This corresponds to: $\boxed{\text{C: } 2x^2+3y-9=0}$

Comparison with stored correct answer

Stored correct answer is C, which matches our derived answer.