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Tangent To a Parabola

'O' is the ve...

Question

$^{'} O^{'}$ is the vertex of the parabola $y^{2} = 4 a x$ and $L$ is the upper end of the latus rectum. If $L H$ is drawn perpendicular to $O L$ meeting $O X$ in $H$ , prove that the length of the double ordinate through $H$ is $4 a \sqrt{5} .$

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Solution

Parabola $: y^{2} = 4 a x$
Ends of saute rectum $(a, \pm 2 a)$
$∴ L (a, 2 a)$
Slope $(O L) = 2$
$L H$ is perpendicular to $O L$
$∴ s l o p e (H L) = \frac{- 1}{2}$
$E q^{n}$ of $H L \Rightarrow (y - 2 a) = \frac{- 1}{2} (y x - a)$
$\Rightarrow + y = (\frac{5 a}{2})$
$x -$ coordinates of $T_{1} = (5 a) . . . .$ [by putting $y = 0$ ]
$T_{1}$ lies on parabola $∴ y^{2} = 4 a (5 a)$
$y = 2 a \sqrt{5}$
Hence, $T_{1} H = 2 a \sqrt{5}$
Thus, Double ordinate $= 4 a \sqrt{5}$

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