JEE Main 2025 — Ellipse Question with Solution

The length of the minor axis is equal to one-fourth of the distance between the foci. Mathematically, this can be expressed as:

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

This simplifies to:

$b=\frac{ae}{4}$

Given that the relationship between $b$, $a$, and $e$ is:

$\frac{b^2}{a^2}=1-e^2$

Substitute $b=\frac{ae}{4}$ into the equation:

${\left(\frac{ae}{4}\right)}^2=a^2\cdot \left(1-e^2\right)$

Expanding and simplifying gives:

$\frac{a^2{e}^2}{16}=a^2(1-e^2)$

Divide both sides by $a^2$:

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

Rearrange and solve for $e^2$:

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

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

Solve for $e$:

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

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

Thus, the eccentricity of the ellipse is:

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