Vertices of Hyperbola

Q. The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are

1. $(6,3)$ and $(-6,3)$
2. $(6,3)$ and $(-2,3)$
3. $(-6,3)$and$(-6,-3)$
4. None of these

Q. The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are

1. $(6,3)$ and $(-6,3)$
2. $(6,3)$ and $(-2,3)$
3. $(-6,3)$and$(-6,-3)$
4. None of these

Q.

The vertices of a hyperbola are (2, 0), (–2, 0) and the foci are (3, 0), (–3, 0). The equation of the hyperbola is

1. $\frac{x^2}{5}-\frac{y^2}{4}=1$

2. $\frac{x^2}{4}-\frac{y^2}{5}=1$

3. $\frac{x^2}{5}-\frac{y^2}{2}=1$

4. $\frac{x^2}{2}-\frac{y^2}{5}=1$

Q. The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are:

1. (6, 3)
2. (-6, 3)
3. (2, 3)
4. (-2, 3)

Q. The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are:

1. (6, 3)
2. (-6, 3)
3. (2, 3)
4. (-2, 3)

Q. The equation of the hyperbola with vertices (3, 0), (–3, 0) and semi-latus rectum 4, is given by

1. $4x^2-3y^2+36=0$
2. $4x^2-3y^2+12=0$
3. $4x^2-3y^2-36=0$
4. None of these