WBJEE · Maths · Ellipse
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
- A \(\frac{1}{4}(\sqrt{5}-1)\)
- B \(\frac{1}{2}(\sqrt{5}+1)\)
- C \(\frac{1}{2}(\sqrt{5}-1)\)
- D \(\frac{1}{4}(\sqrt{5}+1)\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}(\sqrt{5}-1)\)
Step-by-step Solution
Detailed explanation
Given, In ellipse the distance between the foci = Length of the latusrectum \(\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}\)…
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