If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then…
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :
- $\frac{4}{\sqrt{17}}$
- $\frac{\sqrt{3}}{16}$
- $\frac{3}{\sqrt{19}}$
- $\frac{\sqrt{5}}{7}$
Solution
$\begin{aligned} & 2 \mathrm{~b}=\frac{1}{4}(2 \mathrm{ae}) \\ & \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{e}}{4} \\ & \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}} \\ & \mathrm{e}=\sqrt{1-\frac{\mathrm{e}^2}{16}} \\ & \mathrm{e}^2\left(1+\frac{1}{16}\right)=1 \\ & \mathrm{e}=\frac{4}{\sqrt{17}}\end{aligned}$
Asked in: JEE Main 2025 (02 Apr Shift 2)
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