If point P is $(a t^{2}, 2 a t)$ ,then the length of the focal chord PQ is
$a {(t + \frac{1}{t})}^{2}$ .

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Solution

$P (a t^{2}, 2 a t_{1}) . Q (a t^{2}, 2 a t_{2})$
Slope
$t_{1} t_{2} = - 1 \Rightarrow t_{2} = - \frac{1}{t_{1}}$
$P (a t^{2}, 2 a t)$
$t_{1} = t_{2}$
$t_{2} = - \frac{1}{t}$
$Q (a {(- \frac{1}{t})}^{2}, 2 a (- \frac{1}{t})$
$\Rightarrow Q (\frac{a}{t_{2}}, \frac{2 a}{t})$
$P S = \sqrt{{(a t^{2} - a)}^{2} + {(2 a t - a)}^{2}}$
$= \sqrt{a^{2 + 4} + a^{2} + 2 a^{2} t^{2}}$
$= \sqrt{{(a t^{2} + a)}^{2}}$ \
$P S = a t^{2} + a$
$S Q = \frac{a}{t^{2}} + a$
$P Q = P S + S Q = a (t^{2} + 1 + \frac{1}{t^{2}} + 1)$
$= a {(t + \frac{1}{t})}^{2} .$

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