AP EAMCET · Maths · Ellipse
Let \(P, Q\) be the foci of an ellipse and let \(R\) be one end of its minor axis. If \(\triangle P Q R\) is an equilateral triangle, then the eccentricity of the ellipse is equal to
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{8}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(P = (-c, 0), Q = (c, 0), R = (0, b)\) \(PQ = 2c\) \(PR = \sqrt{c^2 + b^2}\) \(2c = \sqrt{c^2 + b^2}\) \(4c^2 = c^2 + b^2\) \(3c^2 = b^2\) \(3c^2 = a^2 - c^2\) \(4c^2 = a^2\) \(e = \frac{c}{a} = \frac{1}{2}\)
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