$P N P^{'}$ is a double ordinate of the parabola ; prove that the locus of the point of intersection of the normal at $P$ and the diameter through P' is the equal parabola $y^{2} = 4 a (x - 4 a) .$

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Solution

Let $P$ be $(a t^{2}, 2 a t)$ then $P^{'}$ is $(a t^{2}, - 2 a t)$

Equation of normal at $P$ is

$y = - t x + 2 a t + a t^{3} . . . . . (i)$

General equation of diameter is $y = \frac{2 a}{m}$

It passes through $P^{'}$

$- 2 a t = \frac{2 a}{m} \Rightarrow m = - \frac{1}{t} \Rightarrow y = - 2 a t . . . . . . (i i)$

Substituting $y$ in $(i)$

$- 2 a t = - t x + 2 a t + a t^{3} t x = 4 a t + a t^{3} x = 4 a + a t^{2}$

Substituting $t$ from $(i i)$

$x = 4 a + a {(\frac{y}{- 2 a})}^{2} x - 4 a = \frac{y^{2}}{4 a} y^{2} = 4 a (x - 4 a)$

Hence proved.

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PNP′ is a double ordinate of the parabola ; prove that the locus of the point of intersection of the normal at P and the diameter through P' is the equal parabola y2=4a(x−4a).

$P N P^{'}$ is a double ordinate of the parabola ; prove that the locus of the point of intersection of the normal at $P$ and the diameter through P' is the equal parabola $y^{2} = 4 a (x - 4 a) .$