Let A be the point (0,4) in the xy-plane and | vedclass

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(B) The coordinates of $A$ are $(0,4)$ and $B$ are $(2t, 0)$.The midpoint $L$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{4+0}{2}\right) = (t, 2)$.The slo

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a parabola

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(B) The coordinates of $A$ are $(0,4)$ and $B$ are $(2t, 0)$.The midpoint $L$ of $AB$ is $\left(\frac{0+2t}{2}, \frac{4+0}{2}\right) = (t, 2)$.The slope of $AB$ is $m_{AB} = \frac{0-4}{2t-0} = -\frac{2}{t}$.The slope of the perpendicular bisector of $AB$ is $m_{\perp} = \frac{t}{2}$.The equation of the perpendicular bisector passing through $L(t, 2)$ is $y - 2 = \frac{t}{2}(x - t)$.To find $M$, set $x = 0$: $y - 2 = \frac{t}{2}(0 - t) \Rightarrow y = 2 - \frac{t^2}{2} = \frac{4-t^2}{2}$.So, $M$ is $\left(0, \frac{4-t^2}{2}\right)$.Let $N(h, k)$ be the midpoint of $LM$. Then $h = \frac{t+0}{2} = \frac{t}{2} \Rightarrow t = 2h$.$k = \frac{2 + \frac{4-t^2}{2}}{2} = \frac{4+4-t^2}{4} = \frac{8-t^2}{4} = 2 - \frac{t^2}{4}$.Substituting $t = 2h$ into the equation for $k$: $k = 2 - \frac{(2h)^2}{4} = 2 - \frac{4h^2}{4} = 2 - h^2$.Thus, the locus of $N(x, y)$ is $y = 2 - x^2$, which represents a parabola.

Let $A$ be the point $(0,4)$ in the $xy$-plane and let $B$ be the point $(2t, 0)$. Let $L$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $M$. Let $N$ be the midpoint of $LM$. Then the locus of $N$ is

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