JEE Main 2025 — Ellipse Question with Solution

To find the length of the latus rectum of the ellipse, we first recognize that the foci of the ellipse are given as $F_1:(2,5)$ and $F_2:(2,-3)$. This indicates that the major axis is aligned along the $y$-axis.

Calculate the distance between the foci:

$F_1{F}_2=8$

The formula involving the distance between the foci and the eccentricity is:

$F_1{F}_2=2be$

Here, the eccentricity $e$ is given as $\frac{4}{5}$. Thus, we can solve for $b$:

$8=2b\cdot \frac{4}{5}⟹b=\frac{8}{2\times \frac{4}{5}}=5$

Determine $a^2$:

Using the relationship between eccentricity, semi-minor axis $b$, and semi-major axis $a$:

$e^2=1-\frac{a^2}{b^2}=1-\frac{a^2}{25}=\frac{16}{25}$

Solving for $a^2$:

$1-\frac{a^2}{25}=\frac{16}{25}⟹\frac{a^2}{25}=\frac{9}{25}⟹a^2=9$

Thus, $a=3$.

Compute the length of the latus rectum:

The formula for the length of the latus rectum $L$ is:

$L=\frac{2a^2}{b}$

Substituting the known values:

$L=\frac{2\times (9)}{5}=\frac{18}{5}$

Therefore, the length of the latus rectum of the ellipse is $\frac{18}{5}$.