The locus of the midpoint of the line segment | vedclass

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Question

(C) Let the moving point on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The focus of the parabola is $S(a, 0)$.Let $(h, k)$ be the midpoint of the line

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$x = 0$

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(C) Let the moving point on the parabola $y^2 = 4ax$ be $(at^2, 2at)$.The focus of the parabola is $S(a, 0)$.Let $(h, k)$ be the midpoint of the line segment joining $(at^2, 2at)$ and $(a, 0)$.Then $h = \frac{at^2 + a}{2}$ and $k = \frac{2at + 0}{2} = at$.From $k = at$,we have $t = \frac{k}{a}$.Substituting $t$ into the expression for $h$: $h = \frac{a(\frac{k}{a})^2 + a}{2} = \frac{\frac{k^2}{a} + a}{2} = \frac{k^2 + a^2}{2a}$.Thus,$2ah = k^2 + a^2$,which simplifies to $k^2 = 2a(h - \frac{a}{2})$.The locus is $y^2 = 2a(x - \frac{a}{2})$.Comparing this with the standard form $Y^2 = 4AX$,where $Y = y$,$X = x - \frac{a}{2}$,and $4A = 2a$ (so $A = \frac{a}{2}$).The directrix is given by $X = -A$,which means $x - \frac{a}{2} = -\frac{a}{2}$.Therefore,$x = 0$.

The locus of the midpoint of the line segment joining the focus to a moving point on the parabola $y^2 = 4ax$ is another parabola with the directrix

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