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Distance Formula

AB , AC are t...

Question

AB, AC 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 tangent to the parabola, then

are in G.P

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are in G.P

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are in A.P

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Solution

The correct option is B are in G.P
Let $B (a t_{1}^{2}, 2 a t_{1}) C (a t_{2}^{2}, 2 a t_{2})$
Since AB, AC are tangents to the parabola we have $A [a t_{1} t_{2}, a (t_{1} + t_{2})]$
Eqution of a tangent to $y^{2} = 4 a x$ is $m x + \frac{a}{m} . . . . . . . . (1)$
$l_{2}$ =Length of the $⊥$ lr from B to (1) = $\frac{m a t_{1}^{2} - 2 a t_{1} + \frac{a}{m}}{\sqrt{1 + m^{2}}}$
Similarly $l_{3}$ = $\frac{m a t_{2}^{2} - 2 a t_{2} + \frac{a}{m}}{\sqrt{1 + m^{2}}}$
$l_{2}$ =Length of the $⊥$ lr from A to the line (1)= $\frac{m a t_{1} t_{2} - a (t_{1} + t_{2}) + \frac{a}{m}}{\sqrt{1 + m^{2}}}$
$l_{2} l_{3} = [\frac{m a t_{1}^{2} - 2 a t_{1} + \frac{a}{m}}{\sqrt{1 + m^{2}}}] [\frac{m a t_{2}^{2} - 2 a t_{2} + \frac{a}{m}}{\sqrt{1 + m^{2}}}]$
$\frac{m^{2} a^{2} t_{1}^{2} t_{2}^{2} + a^{2} (t_{2}^{2} + 2 t_{1} t_{2} + t_{2}^{2}) + \frac{a^{2}}{m^{2}} - 2 a^{2} m t_{1} (t_{1} + t_{2}) - \frac{2 a^{2}}{m} (t_{1} + t_{2}) + 2 a^{2} t_{1} t_{2}}{1 + m^{2}}$
$= [\frac{m a t_{1} t_{2} - a (t_{1} + t_{2}) + \frac{a}{m}}{\sqrt{1 + m^{2}}}] = t_{1}^{2}$
$∴ l_{2}, l_{1}, l_{3}$ are in G.P.

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AB, AC are tangents to a parabola y2=4ax If l1,l2.l3 are the lengths of perpendiculars from A, B, C on any tangent to the parabola, then

AB, AC 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 tangent to the parabola, then