JEE Main 2025 (Online) 29th January Evening Shift | Ellipse Question 11 | Mathematics

If $\alpha x+\beta y=109$ is the equation of the chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, whose mid point is $\left(\frac{5}{2}, \frac{1}{2}\right)$. then $\alpha+\beta$ is equal to :

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, $A < B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

$ \frac{24\sqrt{6}}{5} $

$ \frac{18\sqrt{6}}{5} $

$ 6\sqrt{6} $

$ \frac{12\sqrt{6}}{5} $

JEE Main 2025 (Online) 28th January Evening Shift

MCQ (Single Correct Answer)

-1

If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$, then $\alpha$ is :