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Let hyperbola $C : \frac{x^{2}}{1} - \frac{y^{2}}{3} = 1,$ with the left and right focal points being $F_{1}$ and $F_{2}$ respectively. A line $l$ is drawn through $F_{2}$ to intersect the right half of the hyperbola at points P and Q making $F_{1} P Q = 90^{0},$ then the inscribed circle radius of $Δ F_{1} P Q$ is

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$\sqrt{7} - 1$

$\sqrt{7} - \sqrt{3}$

$\sqrt{5} - 1$

$\sqrt{3} - \sqrt{2}$

answer is C.

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Detailed Solution

$F_{1} F_{2} = 2 a e = 2 \sqrt{4} = 4$

$F_{1} P - F_{2} P = 2 a = Q F_{1} - Q F_{2}$

$F_{1} P Q = \frac{π}{2}$

$P F_{1}^{2} + P F_{2}^{2} = F_{1} F_{2}^{2}$

$P F_{1} + P F_{2} = \sqrt{2 (P F_{1}^{2} + P F_{2}^{2}) - {(P F_{1} - P F_{2})}^{2}} = \sqrt{2 (4^{2}) - {(2)}^{2}} = \sqrt{32 - 4} = \sqrt{28} = 2 \sqrt{7}$

Inscribed circle radius of $Δ F_{1} P Q$ is $r = \frac{1}{2} (F_{1} P + P Q - F_{1} Q) = \frac{1}{2} (P F_{1} + P F_{2} + Q F_{2} - F_{1} Q)$

$= \frac{1}{2} (P F_{1} + P F_{2}) + \frac{1}{2} (Q F_{2} - Q F_{1}) = \frac{1}{2} (2 \sqrt{7}) + \frac{1}{2} (- 2) \Rightarrow r = \sqrt{7} - 1$

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