SOLUTION:
We are to find the equation of the hyperbola with vertices at (-4,0) and (4,0) and foci at (-6,0) and (6,0).
The equation is of the form;
[tex]\frac{x^2}{a^2}\text{ - }\frac{y^2}{b^2}\text{ = 1}[/tex][tex]\begin{gathered} \text{Vertex (a) = 4} \\ \text{Focal length (c) = 6} \end{gathered}[/tex][tex]\begin{gathered} c^2=a^2+b^2 \\ 6^2=4^2+b^2 \\ 36=16+b^2 \\ b^2\text{ = 36 - 16} \\ b^2\text{ = 20} \\ b\text{ = }\sqrt[]{20} \end{gathered}[/tex][tex]\begin{gathered} \frac{x^2}{a^2}\text{ - }\frac{y^2}{b^2}\text{ = 1} \\ \\ \frac{x^2}{4^2}\text{ - }\frac{y^2}{(\sqrt[]{20}^{})^2}\text{ = 1} \\ \\ \frac{x^2}{16^{}}\text{ - }\frac{y^2}{20^{}}\text{ = 1} \end{gathered}[/tex]
