## Calculus (3rd Edition)

$$\left(\frac{x}{40/3}\right)^{2}+\left(\frac{y}{50/3}\right)^{2}=1.$$
Since the foci are at $(0,\pm10)$ and $e=3/5$, then the foci are at the y-axis and $b\gt a$ $$\frac{3}{5}=\frac{10}{b}\Longrightarrow b=\frac{40}{3}$$ $$a^2=b^2-c^2=(\frac{50}{3})^2-100\Longrightarrow a=\frac{40}{3}$$ and hence the ellipse is centered at the origin and has the equation $$\left(\frac{x}{40/3}\right)^{2}+\left(\frac{y}{50/3}\right)^{2}=1.$$