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1
COMEDK 2026 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

The foci of a hyperbola are the same as those of the ellipse with equation $9 x^2+16 y^2=144$.

If the length of the transverse axis of this hyperbola is $2 \cos \alpha$, then its equation is:

A

$$ \frac{x^2}{7-\cos ^2 \alpha}-\frac{y^2}{\cos ^2 \alpha}=1 $$

B

$$ \frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{7-\cos ^2 \alpha}=1 $$

C

$$ \frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{7+\cos ^2 \alpha}=1 $$

D

$$ \frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{5-\cos ^2 \alpha}=1 $$

2
COMEDK 2026 Morning Shift
MCQ (Single Correct Answer)
+1
-0

The difference between the distance of any point on the hyperbola from the two foci is $\mathbf{1 6}$ and the eccentricity is $\mathbf{2}$. Then the equation of the hyperbola is

A

$\frac{x^2}{64}-\frac{y^2}{64}=1$

B

$\frac{x^2}{64}-\frac{y^2}{256}=1$

C

$\frac{x^2}{64}-\frac{y^2}{192}=1$

D

$\frac{x^2}{192}-\frac{y^2}{64}=1$

3
COMEDK 2025 Evening Shift
MCQ (Single Correct Answer)
+1
-0
If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the foci of the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is
A
5
B
9
C
1
D
7
4
COMEDK 2025 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0
The length of the latus rectum of a conic $49 y^2-16 x^2=784$ is
A
$\frac{49}{2}$
B
$\frac{49}{\sqrt{2}}$
C
$\frac{7}{\sqrt{2}}$
D
$\frac{7}{2}$
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