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 $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is :

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Ellipse

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 :

Answer: (D)

Since

a < b

the major axis lies along the

y

-axis, so

e^{2} = 1 - \frac{a ^{2}}{b ^{2}} = \frac{5}{9}

, giving

\frac{a ^{2}}{b ^{2}} = \frac{4}{9}

and

a^{2} = \frac{4 b ^{2}}{9}

. Substituting

(4, 3)

into

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

gives

\frac{16 \cdot 9}{4 b ^{2}} + \frac{9}{b ^{2}} = \frac{36 + 9}{b ^{2}} = 1

, so

b^{2} = 45

and

a^{2} = 20

. The length of the latus rectum is

\frac{2 a ^{2}}{b} = \frac{2 \cdot 20}{45} = \frac{40}{3 5} = \frac{8 5}{3}

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option