EllipseeasyPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Ellipse Latus Rectum with a < b: 8√5/3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

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}

correct

Solution

Step 1: Since

a<b

, the major axis is along the

y

-axis, and eccentricity satisfies

e^2=1-\frac{a^2}{b^2}=\frac59\ \Rightarrow\ \frac{a^2}{b^2}=\frac49.\quad(1)

Step 2: The ellipse passes through

(4,3)

\frac{16}{a^2}+\frac{9}{b^2}=1.\quad(2)

Step 3: From (1),

a^2=\dfrac49 b^2

. Substitute into (2):

\frac{16}{\frac49 b^2}+\frac{9}{b^2}=1\ \Rightarrow\ \frac{36}{b^2}+\frac{9}{b^2}=1\ \Rightarrow\ \frac{45}{b^2}=1\ \Rightarrow\ b^2=45,\ a^2=20.

Step 4: For

a<b

, latus rectum

=\dfrac{2a^2}{b}

\frac{2(20)}{\sqrt{45}}=\frac{40}{3\sqrt5}=\frac{40}{3\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{40\sqrt5}{15}=\frac{8\sqrt5}{3}.

Correct answer: (4)

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