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 : [2025]

Question

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 : [2025]

Smart Achievers · Accepted Answer

(4)

Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be the given ellipse with length of minor axis as 2b and distance between foci is 2ae.

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

We know, $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{e^{2}}{16}}$ [From (i)]

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

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

Solution

Q.
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 : [2025]

1 $\frac{3}{\sqrt{19}}$

2 $\frac{\sqrt{5}}{7}$

3 $\frac{\sqrt{3}}{16}$

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

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Solution

Q. 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 : [2025]

1 319

2 57

3 316

4 417