Let A be a fixed point (0,6) and B be a movin | vedclass

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(C) Given $A = (0,6)$ and $B = (2t, 0)$.The mid-point $M$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{6+0}{2}\right) = (t, 3)$.The slope of $AB$ is $m_{AB

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

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(C) Given $A = (0,6)$ and $B = (2t, 0)$.The mid-point $M$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{6+0}{2}\right) = (t, 3)$.The slope of $AB$ is $m_{AB} = \frac{0-6}{2t-0} = \frac{-6}{2t} = -\frac{3}{t}$.The slope of the perpendicular bisector of $AB$ is $m_{\perp} = -\frac{1}{m_{AB}} = \frac{t}{3}$.The equation of the perpendicular bisector passing through $M(t, 3)$ is $(y-3) = \frac{t}{3}(x-t)$.To find $C$,we set $x=0$ in the equation: $y-3 = \frac{t}{3}(0-t) \Rightarrow y = 3 - \frac{t^{2}}{3}$.So,$C = (0, 3 - \frac{t^{2}}{3})$.Let $P(h, k)$ be the mid-point of $MC$. Since $M = (t, 3)$ and $C = (0, 3 - \frac{t^{2}}{3})$:$h = \frac{t+0}{2} = \frac{t}{2} \Rightarrow t = 2h$.$k = \frac{3 + (3 - \frac{t^{2}}{3})}{2} = \frac{6 - \frac{t^{2}}{3}}{2} = 3 - \frac{t^{2}}{6}$.Substituting $t = 2h$ into the equation for $k$:$k = 3 - \frac{(2h)^{2}}{6} = 3 - \frac{4h^{2}}{6} = 3 - \frac{2h^{2}}{3}$.$3k = 9 - 2h^{2} \Rightarrow 2h^{2} + 3k - 9 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $2x^{2} + 3y - 9 = 0$.

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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