JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let P (10, 2√15) be a point on the hyperbola \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) whose foci are S and S'. If the length of its latus rectum is 8, then the square of the area of \( \triangle PSS' \) is equal to:
- A 4200
- B 900
- C 1462
- D 2700
Answer & Solution
Correct Answer
(D) 2700
Step-by-step Solution
Detailed explanation
\(P (10,2 \sqrt{15})\) lies on \(\frac{ x ^2}{ a ^2}-\frac{ y ^2}{b^2}=1\) \(\therefore \frac{100}{ a ^2}-\frac{60}{b^2}=1\)\(\quad\)....(1) ∵ length of latus rectum \(=8\) \(\frac{2 \cdot b^2}{a}=8 \Rightarrow \frac{b^2}{a}=4\)\(\quad\)....(2) From (1) & (2)…
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