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Q.
If $P S Q$ and $P S^{\prime} R$ are two chords of an ellipse through its foci $S$ and $S$ respectively, then $\frac{P S}{S Q}+\frac{P S^{\prime}}{S^{\prime} R}=$
Conic Sections
Solution:
We know that semi latus rectum is harmonic mean of the segments of focal chord.
$\therefore \frac{1}{S P}+\frac{1}{S Q}=\frac{2 a}{b^{2}}$ ... (i)
and $\frac{1}{S^{\prime} P}+\frac{1}{S^{\prime} R}=\frac{2 a}{b^{2}}$ .... (ii)
Also $S P+S^{\prime} P=2 a$ .... (iii)
From (i), $1+\frac{S P}{S Q}=\frac{2 a}{b^{2}} S P$
From (ii), $1+\frac{S^{\prime} P}{S^{\prime} R}=\frac{2 a}{b^{2}} S^{\prime} P$
On adding, we get
$2+\frac{S P}{S Q}+\frac{S^{\prime} P}{S^{\prime} R}=\frac{2 a}{b^{2}}\left(S P+S^{\prime} P\right)$
$\Rightarrow 2+\frac{S P}{S Q}+\frac{S^{\prime} P}{S^{\prime} R}=\frac{4 a^{2}}{b^{2}}=\frac{4}{1-e^{2}}$
$\Rightarrow \frac{S P}{S Q}+\frac{S^{\prime} P}{S^{\prime} R}=\frac{4}{1-e^{2}}-2=2 \frac{1+e^{2}}{1-e^{2}}$