1
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The equation of the chord, of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose mid-point is $(3,1)$ is :

A
$5 x+16 y=31$
B
$48 x+25 y=169$
C
$4 x+122 y=134$
D
$25 x+101 y=176$
2
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

A
$\frac{1-2 \sqrt{2}}{\sqrt{3}}$
B
$\frac{1-\sqrt{3}}{\sqrt{2}}$
C
$\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$
D
$\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$
3
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
$\frac{5}{3} \sqrt{15}$
4
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
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