Find the product of the perpendiculars from foci upon any tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

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$p_{1} p_{2} = b^{2}$

$p_{1} p_{2} = b^{8}$

$p_{1} p_{2} = b^{3}$

$p_{1} p_{2} = b^{6}$

answer is D.

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Detailed Solution

Let $F_{1}$ and $F_{2}$ be two foci of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

Then $F_{1} = (a e, 0)$ and $F_{2} = (- a e, 0)$

The equation of any tangent to the hyperbola is

$\frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1$ …(i)

Let $p_{1}$ and $p_{2}$ be two perpendiculars from foci upon the tangent (i).

$\begin{array}{l} \Rightarrow p_{1} p_{2} = \frac{a^{2} (e^{2} - 1) (e^{2} \sec^{2} θ - 1)}{((e^{2} - 1) \sec^{2} θ + \tan^{2} θ)} \\ \Rightarrow p_{1} p_{2} = \frac{a^{2} (e^{2} - 1) (e^{2} \sec^{2} θ - 1)}{((e^{2} - 1) \sec^{2} θ + (\sec^{2} θ - 1))} \\ \Rightarrow p_{1} p_{2} = \frac{a^{2} (e^{2} - 1) (e^{2} \sec^{2} θ - 1)}{(e^{2} \sec^{2} θ - 1)} \\ \Rightarrow p_{1} p_{2} = a^{2} (e^{2} - 1) \\ \Rightarrow p_{1} p_{2} = b^{2} \end{array}$