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Let $A$ be the point $(0,4)$ and $B$ be a moving point on $x$-axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$-axis at $R$. The locus of the midpoint $P$ of $M R$ is
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The correct answer is:
$y+x^{2}=2$
Let B $=(2 \alpha, 0)$
$\therefore \mathrm{M}=(\alpha, 2)$
$\mathrm{m}_{\mathrm{AB}}=\frac{4}{-2 \alpha}=-\frac{2}{\alpha}$
$\therefore \mathrm{m}_{\mathrm{MR}}=\frac{\alpha}{2}, \therefore \mathrm{MR} \quad \mathrm{y}-2=\frac{\alpha}{2}(\mathrm{x}-\alpha)$
Put, $\mathrm{x}=0 \Rightarrow \mathrm{y}=2-\frac{\alpha^{2}}{2}$
$\therefore \mathrm{R} \equiv\left(0,2-\frac{\alpha^{2}}{2}\right)$
$\therefore \mathrm{P} \equiv\left(\frac{\alpha}{2}, \frac{2-\alpha^{2} / 2+2}{2}\right) \equiv \left(\frac{\alpha}{2}, 2-\frac{\alpha^{2}}{4}\right)$
$\therefore \mathrm{y}=2-\mathrm{x}^{2} \Rightarrow \mathrm{y}+\mathrm{x}^{2}=2$
$\therefore \mathrm{M}=(\alpha, 2)$
$\mathrm{m}_{\mathrm{AB}}=\frac{4}{-2 \alpha}=-\frac{2}{\alpha}$
$\therefore \mathrm{m}_{\mathrm{MR}}=\frac{\alpha}{2}, \therefore \mathrm{MR} \quad \mathrm{y}-2=\frac{\alpha}{2}(\mathrm{x}-\alpha)$
Put, $\mathrm{x}=0 \Rightarrow \mathrm{y}=2-\frac{\alpha^{2}}{2}$
$\therefore \mathrm{R} \equiv\left(0,2-\frac{\alpha^{2}}{2}\right)$
$\therefore \mathrm{P} \equiv\left(\frac{\alpha}{2}, \frac{2-\alpha^{2} / 2+2}{2}\right) \equiv \left(\frac{\alpha}{2}, 2-\frac{\alpha^{2}}{4}\right)$
$\therefore \mathrm{y}=2-\mathrm{x}^{2} \Rightarrow \mathrm{y}+\mathrm{x}^{2}=2$
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