## Calculus (3rd Edition)

$$\left(\frac{x}{10}\right)^{2}+\left(\frac{y}{8}\right)^{2}=1.$$
Since the vertices are at $(\pm10,0)$ and the foci are at $(\pm6, 0)$, then the foci are at the x-axis $$a=10, \quad b^2=a^2-c^2=100-36=64$$ and hence the ellipse is centered at the origin and has the equation $$\left(\frac{x}{10}\right)^{2}+\left(\frac{y}{8}\right)^{2}=1.$$