JEE Main 2025 (Online) 3rd April Evening Shift | Ellipse Question 7 | Mathematics

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

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 :

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

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

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