JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q10

← JEE Main 2026 · 2 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 10 of 75

Let an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , $a<b$ , pass through the point $(4,3)$ and have eccentricity $\dfrac{\sqrt5}3$ . Then the length of its latus rectum is :

\dfrac{4\sqrt5}3

2\sqrt5

\dfrac{7\sqrt5}3

\dfrac{8\sqrt5}3

Answer and worked solution

Official answer: D

Major axis along y ( $a<b$ ): $e^2=1-\dfrac{a^2}{b^2}=\dfrac59$ , so $a^2=\dfrac49b^2$ .

Through (4, 3). $\dfrac{16}{a^2}+\dfrac9{b^2}=\dfrac{36+9}{b^2}=1$ , so $b^2=45$ and $a^2=20$ .

Latus rectum. $\dfrac{2a^2}b=\dfrac{40}{3\sqrt5}=\dfrac{8\sqrt5}3$ .

(D)

Topics:Ellipse

← Previous question Next question →

JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 10