The locus of the mid-point of the line segmen | vedclass

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(D) Let $Q(h, k)$ be the mid-point of the line segment joining the focus $F(a, 0)$ and a variable point $P(x_0, y_0)$ on the parabola $y^2=4ax$.The co

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$x=0$

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(D) Let $Q(h, k)$ be the mid-point of the line segment joining the focus $F(a, 0)$ and a variable point $P(x_0, y_0)$ on the parabola $y^2=4ax$.The coordinates of the mid-point $Q$ are given by:$(h, k) = \left(\frac{x_0+a}{2}, \frac{y_0+0}{2}\right) = \left(\frac{x_0+a}{2}, \frac{y_0}{2}\right)$From this,we get:$h = \frac{x_0+a}{2} \Rightarrow x_0 = 2h - a$$k = \frac{y_0}{2} \Rightarrow y_0 = 2k$Since $P(x_0, y_0)$ lies on the parabola $y^2=4ax$,we substitute the values of $x_0$ and $y_0$ into the equation:$(2k)^2 = 4a(2h - a)$$4k^2 = 8ah - 4a^2$$k^2 = 2a(h - \frac{a}{2})$Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2a(x - \frac{a}{2})$.This is a parabola of the form $Y^2 = 4AX$,where $Y = y$,$X = x - \frac{a}{2}$,and $4A = 2a \Rightarrow A = \frac{a}{2}$.The equation of the directrix for $Y^2 = 4AX$ is $X = -A$.Substituting the values,we get:$x - \frac{a}{2} = -\frac{a}{2}$$x = 0$

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^2=4ax$ is a conic. The equation of the directrix of that conic is

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