JEE Main 2026 — Apr 2, Shift 1 — Mathematics solution

JEE Main2026Apr 2, Shift 1Mathematics

Let an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , $a < b$ , pass through the point $(4, 3)$ and have eccentricity $\frac{5}{3}$ . Then the length of its latus rectum is :

A
$\frac{4 5}{3}$
B
$25$
C
$\frac{7 5}{3}$
D
$\frac{8 5}{3}$

Solution

Since $a < b$ , the major axis is along $y$ . For this orientation $e^{2} = 1 - a^{2} / b^{2}$ :

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

The point $(4, 3)$ lies on the ellipse: $\frac{16}{a ^{2}} + \frac{9}{b ^{2}} = 1 (2)$ .

From (1), $a^{2} = 4 b^{2} /9$ . Substituting into (2):

$\frac{16 \cdot 9}{4 b ^{2}} + \frac{9}{b ^{2}} = 1 \Rightarrow \frac{36 + 9}{b ^{2}} = 1 \Rightarrow b^{2} = 45, a^{2} = 20$ .

Length of latus rectum $= \frac{2 a ^{2}}{b} = \frac{40}{45} = \frac{40}{3 5} = \frac{8 5}{3}$ .

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