COMEDK 2026 Morning Shift | Hyperbola Question 2 | Mathematics

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

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

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

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

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

COMEDK 2025 Evening Shift

MCQ (Single Correct Answer)

-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

COMEDK 2025 Afternoon Shift

MCQ (Single Correct Answer)

-0

The length of the latus rectum of a conic $49 y^2-16 x^2=784$ is

$\frac{49}{2}$

$\frac{49}{\sqrt{2}}$

$\frac{7}{\sqrt{2}}$

$\frac{7}{2}$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

If the distance between the foci and the distance between the two directrixes are in the ratio $$3: 2$$ for a hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$, then a : b is

$$1: 2$$

$$\sqrt{3}: \sqrt{2}$$

$$2: 1$$

$$\sqrt{2}: 1$$

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