1
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The length of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$, whose mid-point is $\left(1, \frac{1}{2}\right)$, is :

A
$\frac{2}{3} \sqrt{15}$
B
$\frac{1}{3} \sqrt{15}$
C
$\sqrt{15}$
D
$\frac{5}{3} \sqrt{15}$
2
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ and $\mathrm{H}: \frac{x^2}{\mathrm{~A}^2}-\frac{y^2}{\mathrm{~B}^2}=1$. Let the distance between the foci of E and the foci of $H$ be $2 \sqrt{3}$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to :

A
10
B
7
C
9
D
8
3
JEE Main 2024 (Online) 9th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f(x)=x^2+9, g(x)=\frac{x}{x-9}$$ and $$\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3)$$. If $$\mathrm{e}$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1$$, then $$8 \mathrm{e}^2+l^2$$ is equal to.

A
6
B
12
C
8
D
16
4
JEE Main 2024 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the line $$2 x+3 y-\mathrm{k}=0, \mathrm{k}>0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$\mathrm{A}$$ and $$\mathrm{B}$$, respectively. If the equation of the circle having the line segment $$A B$$ as a diameter is $$x^2+y^2-3 x-2 y=0$$ and the length of the latus rectum of the ellipse $$x^2+9 y^2=k^2$$ is $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2 \mathrm{~m}+\mathrm{n}$$ is equal to

A
12
B
13
C
11
D
10
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