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AB, AC are ta...

Question

$A B, A C$ are tangents to a parabola $y^{2} = 4 a x$ , if $l_{1}, l_{2}, l_{3}$ are the lengths of perpendiculars from $A, B, C$ on any tangents to the parabola, then

l_{1}, l_{2}, l_{3}

are in GP

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l_{2}, l_{1}, l_{3}

are in GP

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l_{3}, l_{1}, l_{2}

are in GP

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l_{3}, l_{2}, l_{1}

are in GP

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Solution

The correct option is A $l_{2}, l_{1}, l_{3}$ are in GP
Let the coordinates of $B$ and $C$ be $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ respectively.
Then, the coordinates of $A$ are $(a t_{1} t_{2}, a (t_{1} + t_{2}))$ .
The equation of any tangent to $y^{2} = 4 a x$ is $t y = x + a t^{2}$ .
$l_{1} = \frac{a t_{1} t_{2} - a (t_{1} + t_{2}) t + a t^{2}}{\sqrt{1 + t^{2}}}$
$l_{2} = \frac{a t_{1}^{2} - 2 a t_{1} + a t^{2}}{\sqrt{1 + t^{2}}}$
and $l_{3} = \frac{a t_{2}^{2} - 2 a t_{2} + a t^{2}}{\sqrt{1 + t^{2}}}$
Clearly, $l_{2} l_{3} = l_{1}^{2}$
Therefore, $l_{2}, l_{1}, l_{3}$ are in GP.

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