## College Algebra (10th Edition)

$\quad \displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{13}=1$
The center is at the midpoint of the foci, $(0,0)$ The foci, center and vertices lie on the same line, the main axis. Main axis here: $x=0$ (the $y$-axis). Table 3: Major axis vertical (parallel to the y-axis) $\begin{array}{lll} \text{ Foci}&\text{ Vertices}&\text{ Equation}\\ {(h,k+c)}&{(h,k+a)}&{\displaystyle \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\\ {(h,k-c)}&{(h,k-a)}&{a \gt b \gt 0\text{ and }b^{2}=a^{2}-c^{2}}\end{array}$ --- $x$ -intercepts are $\pm 2 \quad \Rightarrow \quad b=2$ Foci at $(0,\pm 3) \quad \Rightarrow \quad c=3$ Find $a:$ $a^{2}=b^{2}+c^{2}=13$ $a=\sqrt{13}\approx 3.606$ The equation is $\quad \displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{13}=1$