#### Answer

Hyperbola

#### Work Step by Step

Standard form of an hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
In the given equation $-16x^2+36y^2-576=0$,
it can be seen that the coefficients of $x^2$ and $y^2$ are having opposite sign . Thus, the graph of the given equation corresponds to an hyperbola.