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 x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1, a<ba<b, pass through the point (4,3)(4,3) and have eccentricity 53\dfrac{\sqrt5}{3}. Then the length of its latus rectum is
A453\dfrac{4\sqrt5}{3}
B252\sqrt5
C753\dfrac{7\sqrt5}{3}
D853\dfrac{8\sqrt5}{3}correct
Solution
Step 1: Since a<ba<b, the major axis is along the yy-axis, and eccentricity satisfies
e2=1−a2b2=59 ⇒ a2b2=49.(1)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)(4,3):
16a2+9b2=1.(2)\frac{16}{a^2}+\frac{9}{b^2}=1.\quad(2)
 Step 3: From (1), a2=49b2a^2=\dfrac49 b^2. Substitute into (2):
1649b2+9b2=1 ⇒ 36b2+9b2=1 ⇒ 45b2=1 ⇒ b2=45, a2=20.\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<ba<b, latus rectum =2a2b=\dfrac{2a^2}{b}:
2(20)45=4035=4035⋅55=40515=853.\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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