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If $P(\theta)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $S$ and $S^{\prime}$ are foci of the hyperbola,
then $S P . S^{\prime} P=$
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then $S P . S^{\prime} P=$
Solution:
1131 Upvotes
Verified Answer
The correct answer is:
$a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta$

$S P \cdot S P^{\prime}=$ By Distance formula we get;
$=a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta$
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