Answer
$\dfrac{x^2}{9}-\dfrac{y^2}{25}=1$
Work Step by Step
Center: $(0,0)$ and vertices: $(-3,0); (3,0)$ and Asymptote: $ y=\dfrac{5}{3} x $
And we have a horizontal traverse axis.
Standard Equation for a hyperbola is : $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
We have $ a^2=9; b^2 =25$
So, the equation for the hyperbola is:
$\dfrac{x^2}{9}-\dfrac{y^2}{25}=1$
