$A B, A C$ are tangents to a parabola $y^{2} = 4 a x . p_{1}, p_{2}$ & $p_{3}$ are the length of the perpendiculars from $A, B$ & $C$ respectively on any tangent to the curve, then $p_{2}, p_{1}, p_{3}$ are in:

A . P .

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G . P

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H . P

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n o n e o f t h e s e

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Solution

The correct option is B $G . P$
$y^{2} = 4 a x$
Let $B (a t_{1}^{2}, 2 a t_{1}); c = (a t_{2}^{2}, 2 a t_{2})$
So $A = (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Eg of tangent: $y = m x + \frac{a}{m}$
Length of perpendicular from B
$P_{2} = ∣ ∣ ∣ ∣ ∣ \frac{2 a t_{1} - m a t_{1}^{2} - \frac{a}{m}}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ \frac{a}{m} ∣ ∣ \frac{(m t_{1} - 1)^{2}}{\sqrt{1 + m^{2}}}$
Length of perpendicular from C.
$P_{3} = ∣ ∣ ∣ ∣ ∣ \frac{2 a t_{2} - m a t_{2}^{2} - \frac{a}{m}}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ \frac{a}{m} ∣ ∣ \frac{(m t_{2} - 1)^{2}}{\sqrt{1 + m^{2}}}$
Length of perpendicular from A
$P_{1} = ∣ ∣ ∣ ∣ ∣ \frac{m a t_{1} t_{2} - a (t_{1} + t_{2}) + \frac{a}{m}}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ \frac{a}{m} ∣ ∣ \frac{(m t_{1} - 1) (m t_{2} - 1)}{\sqrt{1 + m^{2}}}$
$P_{1}^{2} = P_{2} P_{3}$
So $P_{2}, P_{1}, P_{3}$ are in G.P.

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