JEE Main 2025 Apr 2 Shift 2, Mathematics Q5: Introduction to Ellipse

JEE Main2025Apr 2, Shift 2Mathematics

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 :

Solution

Minor-axis length $= 2 b$ ; distance between foci $= 2 a e$ . The given condition is:

$2 b = \frac{1}{4} (2 a e) \Rightarrow \frac{b}{a} = \frac{e}{4}$ .

Use $b^{2} = a^{2} (1 - e^{2})$ , i.e. $\frac{b ^{2}}{a ^{2}} = 1 - e^{2}$ :

$\frac{e ^{2}}{16} = 1 - e^{2} \Rightarrow e^{2} (1 + \frac{1}{16}) = 1 \Rightarrow e^{2} = \frac{16}{17}$ .

$∴ e = \frac{4}{17}$ .

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