Coordinate Geometry – Ellipse – JEE Main 02 April 2025 Shift 2

Question ID: #1001
JEE Main2 April Shift 2, 2025Coordinate Geometry

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:

  • (1) $\frac{4}{\sqrt{17}}$
  • (2) $\frac{\sqrt{3}}{16}$
  • (3) $\frac{3}{\sqrt{19}}$
  • (4) $\frac{\sqrt{5}}{7}$

Solution:


The eccentricity $e$ of an ellipse is given by the formula:
$$e = \sqrt{1 – \frac{b^2}{a^2}} \Rightarrow \frac{b^2}{a^2} = 1 – e^2$$

$$2b = \frac{1}{4}(2ae)$$

$$b = \frac{ae}{4} \Rightarrow \frac{b}{a} = \frac{e}{4}$$

$$\left(\frac{b}{a}\right)^2 = \frac{e^2}{16}$$

$$1 – e^2 = \frac{e^2}{16}$$

$$1 = e^2 + \frac{e^2}{16}$$

$$1 = \frac{17e^2}{16}$$

$$e^2 = \frac{16}{17}$$

$$e = \frac{4}{\sqrt{17}}$$

Ans. (1)

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