JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(S\) and \(S^{\prime}\) be the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) and \(P (\alpha, \beta)\) be a point on the ellipse in the first quadrant. If \(( SP )^2+\left( S ^{\prime} P \right)^2- SP \cdot S ^{\prime} P =37\), then \(\alpha^2+\beta^2\) is equal to :
- A 15
- B 11
- C 17
- D 13
Answer & Solution
Correct Answer
(D) 13
Step-by-step Solution
Detailed explanation
\(\because P\) lies on ellipse \(\Rightarrow \frac{\alpha^2}{25}+\frac{\beta^2}{9}=1\) \(\because P S+P S^{\prime}=2 a \Rightarrow P S+P S^{\prime}=10\) \(\therefore( PS )^2+\left( PS ^{\prime}\right)^2- PS \cdot PS ^{\prime}=37\)…
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