JEE Main 2025 (Online) 3rd April Morning Shift | Ellipse Question 8 | Mathematics

A line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36}+\frac{y^2}{25}=1$ at $A$ and $B$ such that $(P A) \cdot(P B)$ is maximum. Then $5\left(P A^2+P B^2\right)$ is equal to :

290

377

338

218

JEE Main 2025 (Online) 2nd April Evening Shift

MCQ (Single Correct Answer)

-1

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

$\frac{3}{\sqrt{19}}$

$\frac{\sqrt{3}}{16}$

$\frac{4}{\sqrt{17}}$

$\frac{\sqrt{5}}{7}$

JEE Main 2025 (Online) 2nd April Morning Shift

MCQ (Single Correct Answer)

-1

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and P be a point on the ellipse, then $\min \left(S P \cdot S^{\prime} P\right)+\max \left(S P \cdot S^{\prime} P\right)$ is equal to :

$3(6+\sqrt{2})$

$3(1+\sqrt{2})$

JEE Main 2025 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

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 :