JEE Main 2025 (Online) 4th April Evening Shift | Ellipse Question 5 | Mathematics

The centre of a circle C is at the centre of the ellipse $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$. Let C pass through the foci $F_1$ and $F_2$ of E such that the circle $C$ and the ellipse $E$ intersect at four points. Let P be one of these four points. If the area of the triangle $\mathrm{PF}_1 \mathrm{~F}_2$ is 30 and the length of the major axis of $E$ is 17 , then the distance between the foci of $E$ is :

$\frac{13}{2}$

JEE Main 2025 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

The length of the latus-rectum of the ellipse, whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$, is

$\frac{50}{3}$

$\frac{18}{5}$

$\frac{6}{5}$

$\frac{10}{3}$

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

Let $C$ be the circle of minimum area enclosing the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $\frac{1}{2}$ and foci $( \pm 2,0)$. Let $P Q R$ be a variable triangle, whose vertex $P$ is on the circle $C$ and the side $Q R$ of length $2 a$ is parallel to the major axis of $E$ and contains the point of intersection of $E$ with the negative $y$-axis. Then the maximum area of the triangle $P Q R$ is :

$8(3+\sqrt{2})$

$8(2+\sqrt{3})$

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

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

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

-1

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

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