Question ID: #1001
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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