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Find the locus of the mid-points of chords of the parabola $y^{2} = 4 a x$ which subtends a right angle at the vertex is

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

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

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

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

answer is D.

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Detailed Solution

Let $P = (a t_{1}^{2}, 2 a t_{1}) and Q = (a t_{2}^{2}, 2 a t_{2})$

Slope of $P Q = \frac{2 a (t_{2} - t_{1})}{a (t_{2}^{2} - t_{1}^{2})} = \frac{2}{(t_{1} + t_{2})}$

Equation of PQ is

$\begin{matrix} y - 2 a t_{1} = \frac{2}{(t_{1} + t_{2})} (x - a t_{1}^{2}) \\ \Rightarrow 2 x - (t_{1} + t_{2}) y = - 2 a t_{1} t_{2} \end{matrix}$

Now, $m_{1} = m (O P) = \frac{2}{t_{1}}$

and $m_{2} = m (O Q) = \frac{2}{t_{2}}$

As OP is perpendicular to $O Q$ so

$\begin{aligned} m_{1} m_{2} = - 1 \\ \Rightarrow \frac{4}{t_{1} t_{2}} = - 1 \\ \Rightarrow t_{1} t_{2} = - 4 \end{aligned}$

From Eqs (i) and (ii), we get

$2 (x - 4 a) - (t_{1} + t_{2}) y = 0$

Let M(h, k) be the mid-point of PQ. So

$\begin{aligned} h = \frac{a}{2} (t_{1}^{2} + t_{2}^{2}) and k = \frac{a}{2} (2 t_{1} + 2 t_{2}) \\ \Rightarrow h = \frac{a}{2} (t_{1}^{2} + t_{2}^{2}) and k = a (t_{1} + t_{2}) \end{aligned}$

Now, $k^{2} = a^{2} {(t_{1} + t_{2})}^{2}$

$\begin{array}{l} \Rightarrow k^{2} = a^{2} \{(t_{1}^{2} + t_{2}^{2}) + 2 t_{1} t_{2}\} \\ \Rightarrow k^{2} = a^{2} (2 h + 2 (- 4)) \end{array}$

Thus,

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

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