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WBJEE · Maths · Ellipse

If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

  1. A \(\frac{1}{4}(\sqrt{5}-1)\)
  2. B \(\frac{1}{2}(\sqrt{5}+1)\)
  3. C \(\frac{1}{2}(\sqrt{5}-1)\)
  4. D \(\frac{1}{4}(\sqrt{5}+1)\)
Verified Solution

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 \(\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}\)…