JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let a focus of the ellipse \(E: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) be \(S(4, 0)\) and its eccentricity be \(\dfrac{4}{5}\). If the point \(P(3, \alpha)\) lies on \(E\) and \(O\) is the origin, then the area of \(\triangle POS\) is equal to:
- A \(12/5\)
- B \(14/5\)
- C \(24/5\)
- D \(48/5\)
Answer & Solution
Correct Answer
(C) \(24/5\)
Step-by-step Solution
Detailed explanation
Given the focus of the ellipse \(S(4, 0)\), we have \(ae = 4\). Since the eccentricity \(e = \dfrac{4}{5}\), we get \(a \left(\dfrac{4}{5}\right) = 4 \Rightarrow a = 5\). Using the relation \(b^2 = a^2(1 - e^2)\), we find: \(b^2 = 25 \left(1 - \dfrac{16}{25}\right) = 9\) The…
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