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

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 :

A
$3(6+\sqrt{2})$
B
$3(1+\sqrt{2})$
C
27
D
9
2
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If $\alpha x+\beta y=109$ is the equation of the chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, whose mid point is $\left(\frac{5}{2}, \frac{1}{2}\right)$. then $\alpha+\beta$ is equal to :
A

37

B

46

C

72

D

58

3
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, $A < B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals :

A

$ \frac{24\sqrt{6}}{5} $

B

$ \frac{18\sqrt{6}}{5} $

C

$ 6\sqrt{6} $

D

$ \frac{12\sqrt{6}}{5} $

4
JEE Main 2025 (Online) 28th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If the midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4 / 3)$, and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$, then $\alpha$ is :
A

26

B

18

C

22

D

20

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