Question

Two mutually perpendicular tangents of the parabola $y^{2}=4ax$ at the points $Q_{1}$ and $Q_{2}$ on it meet its axis in $P_{1}$ and $P_{2}.$ If $S$ is the focus of the parabola, then the value of $\left(\frac{1}{S P_{1}} + \frac{1}{S P_{2}}\right)^{- 1}$ is equal to

• A. $\frac{a}{4}$
• B. $\frac{a}{2}$
• C. $a$
• D. $2a$

Answer 1

Answer: (C)

Let $Q_{1}=\left(a t_{1}^{2} , 2 a t_{1}\right)$ & $Q_{2}=\left(a t_{2}^{2} , 2 a t_{2}\right)$ are points of contact of the tangents passing through $P_{1}$ & $P_{2}$ respectively.
$\Rightarrow P_{1}=\left(- a t_{1}^{2} , 0\right)$ & $P_{2}=\left(- a t_{2}^{2} , 0\right)$
$\Rightarrow SP_{1}=a+at_{1}^{2}$ & $SP_{2}=a+at_{2}^{2}$
$\Rightarrow \frac{1}{S P_{1}}+\frac{1}{S P_{2}}=\frac{1}{a + a t_{1}^{2}}+\frac{1}{a + a t_{2}^{2}}=\frac{1}{a + a t_{1}^{2}}+\frac{1}{a + a \left(- \frac{1}{t_{1}}\right)^{2}}$
$=\frac{1}{a}$
$\therefore \left(\frac{1}{S P_{1}} + \frac{1}{S P_{2}}\right)^{- 1}=a$