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

• A. $l_{1}, l_{2}, l_{3}$ are in GP
• B. $l_{2}, l_{1}, l_{3}$ are in GP
• C. $l_{3}, l_{1}, l_{2}$ are in GP
• D. $l_{3}, l_{2}, l_{1}$ are in GP

Answer 1

Answer: (B)

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