Prove that the locus of the middle points of all tangents drawn from points on the directrix to the parabola is
$y^{2} (2 x + a) = a (3 x + a)^{2} .$

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Solution

Consider the parabola $y^{2} = 4 a x$

Equation of directrix is $x = - a$

And tangents are drawn to it on any point $P (a t^{2}, 2 a t)$ then equation of tangent is

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

And the middle point of tangents be $(h, k)$

Put $x = - a$ in $(i)$

$t y = - a + a t^{2} \Rightarrow y = \frac{- a}{t} + a t$

So the point of intersection of tangents with the directrix is $Q (- a, \frac{- a}{t} + a t)$

Mid point of $P Q$ is

$⎛ ⎜ ⎜ ⎝ \frac{a t^{2} - a}{2}, \frac{2 a t - \frac{a}{t} + a t}{2} ⎞ ⎟ ⎟ ⎠ (\frac{a t^{2} - a}{2}, \frac{3 a t^{2} - a}{2 t})$

We considered mid point as $(h, k)$

$\Rightarrow h = \frac{a t^{2} - a}{2} a t^{2} - a = 2 h$
$a t^{2} = 2 h + a$ ..... $(i i)$
$\Rightarrow t = \sqrt{\frac{2 h + a}{a}}$ ..... $(i i i)$

Also $k = \frac{3 a t^{2} - a}{2 t}$

$2 k t = 3 a t^{2} - a$
Substituting $t$ from $(i i i)$ and $a t^{2}$ from $(i i)$ , we get

$2 k \sqrt{\frac{2 h + a}{a}} = 3 (2 h + a) - a k \sqrt{\frac{2 h + a}{a}} = 3 h + a$

Squaring both sides

$k^{2} (\frac{2 h + a}{a}) = {(3 h + a)}^{2} k^{2} (2 h + a) = a {(3 h + a)}^{2}$

Replacing $h$ by $x$ and $k$ by $y$

$y^{2} (2 x + a) = a {(3 x + a)}^{2}$

Hence proved

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