Let A be the point (0,4) and B be a moving po | vedclass

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(A) Let $B = (2\alpha, 0)$.Since $A = (0, 4)$, the midpoint $M$ of $AB$ is $(\alpha, 2)$.The slope of $AB$ is $m_{AB} = \frac{0-4}{2\alpha-0} = -\frac

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$y+x^{2}=2$

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(A) Let $B = (2\alpha, 0)$.Since $A = (0, 4)$, the midpoint $M$ of $AB$ is $(\alpha, 2)$.The slope of $AB$ is $m_{AB} = \frac{0-4}{2\alpha-0} = -\frac{2}{\alpha}$.The slope of the perpendicular bisector of $AB$ is $m_{MR} = -\frac{1}{m_{AB}} = \frac{\alpha}{2}$.The equation of the line $MR$ passing through $M(\alpha, 2)$ with slope $\frac{\alpha}{2}$ is $y-2 = \frac{\alpha}{2}(x-\alpha)$.To find $R$, set $x=0$: $y-2 = \frac{\alpha}{2}(0-\alpha) \Rightarrow y = 2 - \frac{\alpha^{2}}{2}$.So, $R = (0, 2 - \frac{\alpha^{2}}{2})$.Let $P(x, y)$ be the midpoint of $MR$. Then $x = \frac{\alpha+0}{2} = \frac{\alpha}{2}$ and $y = \frac{2 + (2 - \alpha^{2}/2)}{2} = 2 - \frac{\alpha^{2}}{4}$.From $x = \frac{\alpha}{2}$, we have $\alpha = 2x$.Substituting $\alpha = 2x$ into the equation for $y$: $y = 2 - \frac{(2x)^{2}}{4} = 2 - x^{2}$.Thus, $y+x^{2}=2$.

Let $A$ be the point $(0,4)$ and $B$ be a moving point on the $x$-axis. Let $M$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the midpoint $P$ of $MR$ is

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