Mathematics · 2026
JEE Main · 2 April 2026, Shift 1 · Q10
Let an ellipse x^2/(a^2)+y^2/(b^2)=1, a < b, pass through the point (4,3) and have eccentricity (√ 5)/3. Then the length of its latus rectum is:
Let an ellipse $\displaystyle \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}<\mathrm{b}$, pass through the point $\displaystyle (4,3)$ and have eccentricity $\displaystyle \frac{\sqrt{5}}{3}$. Then the length of its latus rectum is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{8 \sqrt{5}}{3}$
More from Ellipse
- Let x=9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 1/3. Let P(α, 0),…2026
- Consider the parabola P: y^2=4 k x and the ellipse E: x^2/a^2+y^2/b^2=1. Let the line segment joining the…2026
- Let x^2/(f(a^2+7 a+3))+y^2/(f(3 a+15))=1 represent an ellipse with major axis along y -axis, where f is a…2026
- Let A be the point (3,0) and circles with variable diameter AB touch the circle x^2+y^2=36 internally. Let…2026
- Let a focus of the ellipse E: x^2/a^2+y^2/b^2=1 be S(4,0) and its eccentricity be 4/5. If the point P(3, α)…2026
JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.