JEE Main 2026 — Ellipse Question with Solution

Given the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a<b$.

The eccentricity is given by $e=\frac{\sqrt{5}}{3}$.

Since $a<b$, the relation between $a$, $b$, and $e$ is $a^2=b^2(1-e^2)$.

Substituting the value of $e$:

$a^2=b^2\left(1-\frac{5}{9}\right)=\frac{4}{9}{b}^2$

The ellipse passes through the point $(4,3)$, so substituting $x=4$ and $y=3$ into the equation of the ellipse gives:

$\frac{16}{a^2}+\frac{9}{b^2}=1$

Substituting $a^2=\frac{4}{9}{b}^2$ into the above equation:

$\frac{16}{\frac{4}{9}{b}^2}+\frac{9}{b^2}=1$

$\frac{36}{b^2}+\frac{9}{b^2}=1$

$\frac{45}{b^2}=1⟹b^2=45$

Now, finding $a^2$:

$a^2=\frac{4}{9}\times 45=20$

For an ellipse with $a<b$, the length of the latus rectum is given by $\frac{2a^2}{b}$.

Substituting the values of $a^2$ and $b=\sqrt{45}=3\sqrt{5}$:

Length of latus rectum $=\frac{2\times 20}{3\sqrt{5}}=\frac{40}{3\sqrt{5}}=\frac{8\sqrt{5}}{3}$

Answer: $\frac{8\sqrt{5}}{3}$