If double ordinate of parabola $y^{2} = 4 a x$ subtends right angle at vertex, then length of double ordinate is double the lactic rectum $(8 a)$ .

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Solution

Given parabola is $y^{2} = 4 a x .$
The coordinates of ends of double ordinate be $(a t_{1}^{2}, 2 a t_{1})$
and $(a t_{2}^{2}, 2 a t_{2})$ . Also $t_{2} = - t_{1}$ (double ordinate $⊥$ lar to x axis)
Double ordinate subtends $90^{\circ}$ at origin.
So, $m_{1} m_{2} = - 1$
$\Rightarrow \frac{2 a t_{1}}{a t_{1}^{2}} \times \frac{2 a t_{2}}{a t_{2}^{2}} = - 1$
$\Rightarrow t_{1} t_{2} = - 4 \Rightarrow t_{1}^{2} = 4, t_{2}^{2} = 4, t_{1} t_{2} = - 4$
Length of the double ordinate $= \sqrt{a^{2} (t_{1}^{2} - t_{2}^{2})^{2} + 4 a^{2} (t_{1} - t_{2})^{2}}$
$= \sqrt{a^{2} (t_{1}^{4} + t_{2}^{4} - 2 t_{1}^{2} t_{2}^{2} + 4 t_{1}^{2} + 4 t_{2}^{2} - 8 t_{1} t_{2})}$
$= \sqrt{a^{2} (2 \times 16 - 2 \times 16 + 4 \times 8 + 8 \times 4)}$
$= 8 a .$

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