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Slope Point Form of a Line

Two mutually ...

Question

Two mutually perpendicular tangent of the parabola $y^{2} = 4 a x$ meet the axis in $P_{1}$ and $P_{2}$ . If $S$ is the focus of the parabola, then $\frac{1}{(S P_{1})} + \frac{1}{(S P_{2})}$ is equal to

\frac{4}{a}

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\frac{2}{a}

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\frac{1}{a}

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\frac{1}{4 a}

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Solution

The correct option is A $\frac{1}{a}$
Let $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ be any two points on the given parabola
Equation of tangent to the parabola $y^{2} = 4 a x$ at $(a t_{1}^{2}, 2 a t_{1})$ is
$t_{1} y = x + a t_{1}^{2}$
Since this tangent meets x-axis in $P_{1}$ .
Coordinates of $P_{1}$ are $(- a t_{1}^{2}, 0)$
Equation of tangent to the parabola $y^{2} = 4 a x$ at $(a t_{2}^{2}, 2 a t_{2})$ is
$t_{2} y = x + a t_{2}^{2}$
Since, this tangent meets x-axis in $P_{2}$ .
Coordinates of $P_{2}$ are $(- a t_{2}^{2}, 0)$
$S P_{1} = a (1 + t_{1}^{2}); S P_{2} = a (1 + t_{2}^{2})$
$\Rightarrow t_{1} t_{2} = - 1$
$\frac{1}{S P_{1}} = \frac{1}{a (1 + t_{1}^{2})}$
$\frac{1}{S P_{2}} = \frac{1}{a (1 + t_{2}^{2})} = \frac{t_{1}^{2}}{a (t_{1}^{2} + 1)}$
$∴ \frac{1}{S P_{1}} + \frac{1}{S P_{2}} = \frac{1}{a}$

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