## Precalculus (6th Edition) Blitzer

From $c^2=a^2-b^2$, divide by $a^2$ on both sides; we have $\frac{c^2}{a^2}=1-\frac{b^2}{a^2}$. As $\frac{c}{a}\to 0$, we have $\frac{b^2}{a^2}\to 1$, which means that the graph will approach to a circle.