A focal chord of the parabola $y^{2} = 4 a x$ meets it at $P$ and $Q$ . If $S$ is the focus then $\frac{1}{S P} + \frac{1}{S Q} =$

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$1 / a$

$a$

$2 a$

$2 / a$

answer is B.

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Detailed Solution

Let $P = (a t_{1}^{2}, 2 a t_{1}), Q = (a t_{2}^{2}, 2 a t_{2})$ . Now focus $S = (a, 0)$

$Since \vec{P Q} is a focal chord, t_{1} t_{2} = - 1$

$\begin{array}{l} S P = \sqrt{{(a t_{1}^{2} - a)}^{2} + {(2 a t_{1} - 0)}^{2}} = a \sqrt{{(t_{1}^{2} - 1)}^{2} + 4 t_{1}^{2}} = a \sqrt{{(t_{1}^{2} + 1)}^{2}} = a (t_{1}^{2} + 1) \\ S Q = \sqrt{{(a t_{2}^{2} - a)}^{2} + {(2 a t_{1} - 0)}^{2}} = a \sqrt{{(t_{2}^{2} - 1)}^{2} + 4 t_{2}^{2}} = a \sqrt{{(t_{2}^{2} + 1)}^{2}} = a (t_{2}^{2} + 1) \\ \frac{1}{S P} + \frac{1}{S Q} = \frac{1}{a (t_{1}^{2} + 1)} + \frac{1}{a (t_{2}^{2} + 1)} = \frac{1}{a} [\frac{1}{t_{1}^{2} + 1} + \frac{1}{t_{2}^{2} + 1}] = \frac{1}{a} [\frac{t_{2}^{2} + 1 + t_{1}^{2} + 1}{(t_{2}^{2} + 1) (t_{2}^{2} + 1)}] \\ = \frac{1}{a} [\frac{t_{1}^{2} + t_{2}^{2} + 2}{t_{1}^{2} t_{2}^{2} + t_{1}^{2} + t_{2}^{2} + 1}] = \frac{1}{a} [\frac{t_{1}^{2} + t_{2}^{2} + 2}{1 + t_{1}^{2} + t_{2}^{2} + 1}] = \frac{1}{a} \end{array}$