Maths Q12 – JEE-MAIN 2025

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 :

Clue 1 of 3: Ellipse Properties

Recall the standard definitions for the length of the minor axis and the distance between the foci of an ellipse.

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Step 1: Define parameters and set up the given condition✦ Active

For an ellipse, let $a$ be the semi-major axis, $b$ be the semi-minor axis, and $e$ be the eccentricity. The length of the minor axis is $2 b$ , and the distance between the foci is $2 a e$ . The given condition is:

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

This simplifies to $4 b = a e$ .

Step 2: Use the fundamental relationship for an ellipse○ Expand

The relationship between $a$ , $b$ , and $e$ for an ellipse is $b^{2} = a^{2} (1 - e^{2})$ . From the simplified condition in Step 1, we have $b = \frac{a e}{4}$ . Substitute this into the fundamental relationship:

(\frac{a e}{4})^{2} = a^{2} (1 - e^{2})

This expands to $\frac{a^{2} e^{2}}{16} = a^{2} (1 - e^{2})$ .

Step 3: Solve for the eccentricity

e

○ Expand

Divide both sides by $a^{2}$ (since $a \neq 0$ ) and solve for $e$ :

\frac{e^{2}}{16} = 1 - e^{2} e^{2} = 16 - 16 e^{2} 17 e^{2} = 16 e^{2} = \frac{16}{17} e = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}}

The eccentricity of the ellipse is $\frac{4}{\sqrt{17}}$ .

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Maths Question 12 – JEE-MAIN 2025

Clue 1 of 3: Ellipse Properties

🧠 Full Solution Path