1
JEE Main 2025 (Online) 7th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let e1 and e2 be the eccentricities of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and the hyperbola $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$, respectively. If b < 5 and e1e2 = 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

A

$\frac{4}{5}$

B

$\frac{3}{5}$

C

$\frac{\sqrt{7}}{4}$

D

$\frac{\sqrt{3}}{2}$

2
JEE Main 2025 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ be $8 \sqrt{\frac{5}{3}}$. If for H , the length of the latus rectum is $l$ and the product of the focal distances of the point P is m , then $9 l^2+6 \mathrm{~m}$ is equal to :

A
187
B
184
C
186
D
185
3
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let one focus of the hyperbola $\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x=\frac{9}{\sqrt{10}}$. If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of H , then $9\left(e^2+l\right)$ is equal to :

A
12
B
14
C
15
D
16
4
JEE Main 2025 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6)$, then the length of its latus-rectum is :

A
$\frac{25}{6}$
B
$\frac{144}{5}$
C
$\frac{288}{5}$
D
$\frac{24}{5}$
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