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If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is
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Verified Answer
The correct answer is:
$\frac{1}{2}(\sqrt{5}-1)$
Given,
In ellipse the distance between the foci
= Length of the latusrectum
$\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}$
$\because \quad e^{2}=1-\frac{b^{2}}{a^{2}}$
$\Rightarrow \quad e^{2}=1-e \quad$ [from Eq.
$\Rightarrow e^{2}+e-1=0$
$\Rightarrow \quad e=\frac{-1 \pm \sqrt{1+4}}{2}$
(by quadratic formula) $\Rightarrow \quad e=\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow$
$$
e=\frac{\sqrt{5}-1}{2} \quad(\because 1>e>0)
$$
In ellipse the distance between the foci
= Length of the latusrectum
$\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}$
$\because \quad e^{2}=1-\frac{b^{2}}{a^{2}}$
$\Rightarrow \quad e^{2}=1-e \quad$ [from Eq.
$\Rightarrow e^{2}+e-1=0$
$\Rightarrow \quad e=\frac{-1 \pm \sqrt{1+4}}{2}$
(by quadratic formula) $\Rightarrow \quad e=\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow$
$$
e=\frac{\sqrt{5}-1}{2} \quad(\because 1>e>0)
$$
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