Question

For hyperbola $-\frac{x^2}{9}+\frac{y^2}{16}=1$, focus is is on

• A. x-axis
• B. y-axis
• C. z-axis
• D. none

Answer 1

Answer: (B)

y-axis

The given hyperbola $\frac{-x^2}{9}+\frac{y^2}{16}=1$ or $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is a standard form of a conjugate hyperbola.

By comparing it with it's standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$

We can know $a=3$ and $b=4$

For a standard form of a conjugate hyperbola, the foci lie on its transverse axis. $i.e.$ $y$ - axis. one each side of the hyperbola, at $(0,be)$ and $(0,-be)$ respectively.