The locus of the middle points of all tangents drawn form points on the directrix to the parabola $y^{2} = 4 a x$ is

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

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

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

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

answer is D.

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Detailed Solution

Let the tangent from a point A on the directrix to the parabola touch it at $P (a t^{2}, 2 a t)$ .
Then equation of tangent at $P (a t^{2}, 2 a t)$ is

Question Image
$t y = x + a t^{2}$ … (1)
Solving (1) and $x + a = 0$ (direct) then we get
$x = - a$ and $y = \frac{- a + a t^{2}}{t}$
$∴$ co-ordinates of A is $(- a, \frac{- a + a t^{2}}{t})$
Let the mid point of AP is $Q (h, k)$ then
   $h = \frac{a t^{2} - a}{2}$
or $t^{2} = \frac{2 h + a}{a}$ …. (2)
and    $k = \frac{2 a t + \frac{- a + a t^{2}}{t}}{2}$
or $2 k t = - a + 3 a t^{2}$
or $4 k^{2} t^{2} = a^{2} {(3 t^{2} - 1)}^{2}$
$4 k^{2} \times (\frac{2 h + a}{a}) = a^{2} {\frac{6 h + 3 a}{a} - 1}^{2}$ (From (2))
or $4 (2 h + a) k^{2} = a {(6 h + 2 a)}^{2}$
or $(2 h + a) k^{2} = a {(3 h + a)}^{2}$
Hence locus of mid point    $Q (h, k)$ is
$y^{2} (2 x + a) = a {(3 x + a)}^{2}$