Prove that the locus of the middle point of the portion of a normal intersected between the curve and the axis is a parabola whose vertex is the focus and whose latus rectum is one quarter of that of the original parabola.

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Solution

Let the equation of parabola be $y^{2} = 4 a x$ and the middle point of intercepted portion be $(h, k)$

Equation of normal at $P (a t^{2}, 2 a t)$ is

$y = - t x + 2 a t + a t^{3}$

Put $y = 0$

$0 = - t x + 2 a t + a t^{3} \Rightarrow x = 2 a + a t^{2}$

So, the point of intersection with axis is $Q (2 a + a t^{2}, 0)$

Mid point of $P Q$ is

$(\frac{a t^{2} + 2 a + a t^{2}}{2}, \frac{2 a t + 0}{2}) (a (1 + t^{2}), a t) a (1 + t^{2}) = h . . . . (i) a t = k \Rightarrow t = \frac{k}{a}$

Substitute $t$ in $(i)$

$a (1 + {(\frac{k}{a})}^{2}) = h k^{2} + a^{2} = a h k^{2} = a (h - a)$

So, the general equation is

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

Length of latus rectum $= 4 \times \frac{a}{4} = a$

which is one quarter the latusrectum of orignal parabola

Vertex of parabola is

$x - a = 0, y = 0 \Rightarrow (a, 0)$

which is focus of orignal parabola

Hence proved.

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