## College Algebra (6th Edition)

Yes, because if I plug in $y=-\frac{2}{3}x$ into the hyperbola's equation I get $\frac{x^2}{9}-\frac{(-\frac{2}{3}x)^2}{4}=\frac{x^2}{9}-\frac{\frac{4}{9}x^2}{4}=\frac{x^2}{9}-\frac{x^2}{9}=0\ne1$, thus they have no points in common