The locus of the middle points of all chords | vedclass

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(C) Let the midpoint of a chord passing through the vertex $(0, 0)$ be $(h, k)$.Since the chord passes through the vertex $(0, 0)$ and the midpoint $(

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a parabola

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(C) Let the midpoint of a chord passing through the vertex $(0, 0)$ be $(h, k)$.Since the chord passes through the vertex $(0, 0)$ and the midpoint $(h, k)$, the equation of the chord is $y - 0 = \frac{k - 0}{h - 0}(x - 0)$, which simplifies to $y = \frac{k}{h}x$.Substituting this into the parabola equation $y^2 = 4ax$, we get $(\frac{k}{h}x)^2 = 4ax$.$\frac{k^2}{h^2}x^2 = 4ax$.Since $x=0$ is the vertex, for the chord, we consider $x 
eq 0$, so $\frac{k^2}{h^2}x = 4a$, which gives $x = \frac{4ah^2}{k^2}$.However, the midpoint $(h, k)$ must satisfy the chord equation $k = \frac{k}{h}h$, which is trivial.The condition for $(h, k)$ to be the midpoint of a chord of $y^2 = 4ax$ is $k^2 = 2a(h - x_1)$ where $x_1$ is the other end. For a chord through the vertex, the equation of the chord with midpoint $(h, k)$ is $T = S_1$, i.e.,$yk - 2a(x + h) = k^2 - 4ah$.Since it passes through $(0, 0)$, we have $0 - 2ah = k^2 - 4ah$, which simplifies to $k^2 = 2ah$.Replacing $(h, k)$ with $(x, y)$, the locus is $y^2 = 2ax$, which is a parabola.

$A$. $P$	$I$. $(2,0)$
$B$. $Q$	$II$. $(\frac{3}{2}, -\frac{1}{4})$
$C$. $S$	$III$. $(\frac{3}{2}, 0)$
$D$. $Z$	$IV$. $(\frac{3}{2}, -\frac{1}{2})$
	$V$. $(0, \frac{3}{2})$

The locus of the middle points of all chords of the parabola $y^2 = 4ax$ passing through the vertex is

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