If a focal chord of the parabola be at a distance $d$ from the vertex, then its length is equal to

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$\frac{a^{2}}{d^{2}}$

$\frac{2 a^{2}}{d}$

$\frac{4 a^{3}}{d^{2}}$

$\frac{d^{2}}{a}$

answer is C.

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Detailed Solution

Chord PQ joining $t_{1}, t_{2}$ is

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

where $t_{1} t_{2} = - 1$ being focal chord

$\begin{array}{l} P Q = P S + S Q = a (1 + t_{1}^{2} + 1 + t_{2}^{2}) \\ = a (2 + t_{1}^{2} + \frac{1}{t_{1}^{2}}) \end{array}$

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

Also $d = distance of P Q$ from origin

$\begin{array}{l} ∴ d = \frac{|- 2 a t_{1} t_{2}|}{\sqrt{{(t_{1} + t_{2})}^{2} + 4}} = \frac{2 a}{\sqrt{{(t_{1} - \frac{1}{t_{1}})}^{2} + 4}} \\ d = \frac{2 a}{t_{1} + \frac{1}{t_{1}}} \Rightarrow t_{1} + \frac{1}{t_{1}} = \frac{2 a}{d} \\ ∴ P Q = a \cdot \frac{4 a^{2}}{d^{2}} = \frac{4 a^{3}}{d^{2}} \Rightarrow (3). \end{array}$