JEE Main 2026 — Mathematics Coordinate Geometry

Given the equation of the ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$a2x2​+b2y2​=1 with $a < b$a<b.

The eccentricity is given by $e = \dfrac{\sqrt{5}}{3}$e=35​​.

Since $a < b$a<b, the relation between $a$a, $b$b, and $e$e is $a^2 = b^2(1 - e^2)$a2=b2(1−e2).

Substituting the value of $e$e:

$$a^2 = b^2\left(1 - \dfrac{5}{9}\right) = \dfrac{4}{9}b^2$$a2=b2(1−95​)=94​b2

The ellipse passes through the point $(4, 3)$(4,3), so substituting $x = 4$x=4 and $y = 3$y=3 into the equation of the ellipse gives:

$$\dfrac{16}{a^2} + \dfrac{9}{b^2} = 1$$a216​+b29​=1

Substituting $a^2 = \dfrac{4}{9}b^2$a2=94​b2 into the above equation:

$$\dfrac{16}{\dfrac{4}{9}b^2} + \dfrac{9}{b^2} = 1$$94​b216​+b29​=1

$$\dfrac{36}{b^2} + \dfrac{9}{b^2} = 1$$b236​+b29​=1

$$\dfrac{45}{b^2} = 1 \implies b^2 = 45$$b245​=1⟹b2=45

Now, finding $a^2$a2:

$$a^2 = \dfrac{4}{9} \times 45 = 20$$a2=94​×45=20

For an ellipse with $a < b$a<b, the length of the latus rectum is given by $\dfrac{2a^2}{b}$b2a2​.

Substituting the values of $a^2$a2 and $b = \sqrt{45} = 3\sqrt{5}$b=45​=35​:

Length of latus rectum $= \dfrac{2 \times 20}{3\sqrt{5}} = \dfrac{40}{3\sqrt{5}} = \dfrac{8\sqrt{5}}{3}$=35​2×20​=35​40​=385​​

Answer: $\dfrac{8\sqrt{5}}{3}$385​​