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.

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

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