The eccentricity of an ellipse whose latus re | vedclass

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Question

(B) For an ellipse with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$,the length of the latus rectum is $\frac{2b^2}{a}$ and the distance

Vedclass · Accepted Answer

$\frac{\sqrt{5} - 1}{2}$

Vedclass · Answer

(B) For an ellipse with equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$,the length of the latus rectum is $\frac{2b^2}{a}$ and the distance between the foci is $2ae$.Given that the latus rectum is equal to the distance between the foci:$\frac{2b^2}{a} = 2ae$$\frac{b^2}{a} = ae$$b^2 = a^2e$Since $b^2 = a^2(1 - e^2)$,we substitute this into the equation:$a^2(1 - e^2) = a^2e$$1 - e^2 = e$$e^2 + e - 1 = 0$Using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$e = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$Since the eccentricity $e$ must be positive and $e $e = \frac{\sqrt{5} - 1}{2}$

The eccentricity of an ellipse whose latus rectum is equal to the distance between its two foci is:

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