Question

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

• A. $(3,3)$
• B. $(5,5)$
• C. $(0,0)$
• D. $(4,5)$

Answer 1

Answer: (C)

$(0,0)$

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 center lies at origin at $(0,0)$. Hence the correct option is $C$.