The locus of middle point of the portion of the normal to $y^{2} = 4 a x$ intercepted between curve and axis of parabola is

y^{2} = a (x - a)

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y^{2} = 2 a (x - a)

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y^{2} = (x - 2 a)

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y^{2} = a^{2} (x - 2 a)

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Solution

The correct option is A $y^{2} = a (x - a)$
Given parabola is $y^{2} = 4 a x$
Equation of normal at $P (a t^{2}, 2 a t)$ is
$t x + y = 2 a t + a t^{3}$
Point of intersection of normal with axis of parabola is
$\Rightarrow Q = (2 a + a t^{2}, 0)$

Let the midpoint be $R (h, k)$ , so
$h = \frac{2 a + a t^{2} + a t^{2}}{2}, k = \frac{2 a t}{2} \Rightarrow h = a + a t^{2}, k = a t \Rightarrow h = a + \frac{a k^{2}}{a^{2}} \Rightarrow a h = a^{2} + k^{2}$

Hence, the required locus is
$y^{2} = a (x - a)$

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Similar questions

y^{2} = 4 a x

The locus of the midpoints of the focal chords of the parabola $y^{2} = 4 a x$ is

y^{2} = 4 a x

A (t_{1})

B (t_{2})

C (t_{3})

y^{2} = 4 a x,

A B

y^{2} = 16 x

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