$If P S Q and P S^{'} R are two chords of an ellipse through its foci S and S respectively, then \frac{P S}{S Q} + \frac{P S^{'}}{S^{'} R} =$

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$2 (\frac{1 + e^{2}}{1 - e^{2}})$

$\frac{1 - e^{2}}{1 + e^{2}}$

$\frac{1 + e^{2}}{1 - e^{2}}$

$2 (\frac{1 - e^{2}}{1 + e^{2}})$

answer is A.

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Detailed Solution

$We know that semi latus rectum is harmonic mean of the segments of focal chord.$

$\begin{aligned} ∴ \frac{1}{S P} + \frac{1}{S Q} = \frac{2 a}{b^{2}} - - - - (i) \\ and \frac{1}{S^{'} P} + \frac{1}{S^{'} R} = \frac{2 a}{b^{2}} - - - (i i) \\ Also S P + S^{'} P = 2 a - - - - (i i i) \\ From (i), 1 + \frac{S P}{S Q} = \frac{2 a}{b^{2}} S P \\ From (ii), 1 + \frac{S^{'} P}{S^{'} R} = \frac{2 a}{b^{2}} S^{'} P \end{aligned}$

$\begin{array}{l} On adding, we get 2 + \frac{S P}{S Q} + \frac{S^{'} P}{S^{'} R} = \frac{2 a}{b^{2}} (S P + S^{'} P) \\ \Rightarrow 2 + \frac{S P}{S Q} + \frac{S^{'} P}{S^{'} R} = \frac{4 a^{2}}{b^{2}} = \frac{4}{1 - e^{2}} \\ \Rightarrow \frac{S P}{S Q} + \frac{S^{'} P}{S^{'} R} = \frac{4}{1 - e^{2}} - 2 = 2 \frac{1 + e^{2}}{1 - e^{2}} \end{array}$