A = (-2, 0) and P is a point on the parabola | vedclass

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(A) Let the point $P$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.Given $A = (-2, 0)$,let $Q = (h, k)$ be the midpoint of $\overline{AP}$.Then $h = \fr

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$(0, 0)$

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(A) Let the point $P$ on the parabola $y^2 = 8x$ be $(2t^2, 4t)$.Given $A = (-2, 0)$,let $Q = (h, k)$ be the midpoint of $\overline{AP}$.Then $h = \frac{2t^2 - 2}{2} = t^2 - 1$ and $k = \frac{4t + 0}{2} = 2t$.From $k = 2t$,we have $t = \frac{k}{2}$.Substituting $t$ into the equation for $h$: $h = (\frac{k}{2})^2 - 1 = \frac{k^2}{4} - 1$.Rearranging gives $k^2 = 4(h + 1)$.The locus of $Q$ is $y^2 = 4(x + 1)$.Comparing this with the standard form $Y^2 = 4aX$,where $Y = y$,$X = x + 1$,and $4a = 4$,we get $a = 1$.The vertex of this parabola is $(-1, 0)$.The focus is at $(X + a, Y) = (-1 + 1, 0) = (0, 0)$.

$A = (-2, 0)$ and $P$ is a point on the parabola $y^2 = 8x$. If $Q$ bisects $\overline{AP}$ and the locus of $Q$ is a parabola,then its focus is

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