Prove that the semi-latus-rectum of a parabola is a harmonic mean between the segments of any focal chord.

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Solution

Let the parabola be $y^{2} = 4 a x$

Let the two ends of focal chord be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$

Focus of the parabola is $S (a, 0)$

$P S = \sqrt{{(a t_{1}^{2} - a)}^{2} + {(2 a t_{1})}^{2}}$
$P S = \sqrt{a^{2} t_{1}^{4} + a^{2} - {2 a}^{2} t_{1}^{2} + 4 a^{2} t_{1}^{2}}$
$P S = \sqrt{a^{2} t_{1}^{4} + a^{2} {+ 2 a}^{2} t_{1}^{2}} = \sqrt{{(a t_{1}^{2} + a)}^{2}} = a t_{1}^{2} + a$
$P S = a (t_{1}^{2} + 1)$

Similarly $Q S = a (t_{2}^{2} + 1)$

Harmonic mean between $P S$ and $Q S$ is

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

For focal chord $t_{1} t_{2} = - 1$

$H = 2 a \frac{1 + t_{1}^{2} + t_{2}^{2} + 1}{t_{1}^{2} + t_{2}^{2} + 2}$
$H = 2 a \frac{t_{1}^{2} + t_{2}^{2} + 2}{t_{1}^{2} + t_{2}^{2} + 2} = 2 a$

$H = 2 a$

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