1
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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 :
A
$8(3+\sqrt{2})$
B
$8(2+\sqrt{3})$
C
$6(3+\sqrt{2})$
D
$6(2+\sqrt{3})$
2
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The shortest distance between the curves $y^2=8 x$ and $x^2+y^2+12 y+35=0$ is:
A
$2 \sqrt{3}-1$
B
$2 \sqrt{2}-1$
C
$3 \sqrt{2}-1$
D
$\sqrt{2}$
3
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider the lines $x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5, \lambda$ being a parameter, all passing through a point P. One of these lines (say $L$ ) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$, then the value of $d^2$ is
A
10
B
20
C
15
D
30
4
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $y=y(x)$ be the solution of the differential equation

$\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

A
$\frac{4}{3}$
B
$\frac{2}{3}+e^3$
C
$\frac{4}{3}+e^3$
D
$\frac{2}{3}$
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