Show that the normal at a point ( ${a t}^{2} .2 a t$ ) to the parabola meets the curve again at the point ( ${a f}^{2} .2 a f$ ). Where $f = - t - \frac{2}{t}$ .

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Solution

$y^{2} = 4 a x$ ........ $(1)$
$2 y \frac{d y}{d x} = 4 a$
$\Rightarrow \frac{d y}{d x} = \frac{2 a}{y}$
At the point $(a t^{2}, 2 a t)$
$\frac{d y}{d x} = \frac{2 a}{2 a t} = \frac{1}{t}$
Equation of the normal at the point $(a t^{2}, 2 a t)$ is
$(t - 2 a t) (\frac{1}{t}) + x - a t^{2} = 0$
$\Rightarrow y + x t = 2 a t + a t^{3}$ ........ $(2)$
If normal point again meets at
$(a f^{2}, 2 a f)$ then this point will satisfy $(2)$
$\Rightarrow 2 a f + a f t = 2 a t + a t^{3}$
$\Rightarrow a (f - t) [2 + t (f + t)] = 0$
$\Rightarrow 2 + t (f + t) = 0$
$∴ t (f + t) = - 2$
$∴ f + t = \frac{- 2}{t}$
$∴ f = - t - \frac{2}{t}$

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