College Algebra, Chapter 8, 8.2, Section 8.2, Problem 28

Determine the equation of the ellipse whose graph is given below.


The form $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ is an ellipse with vertical major axis and whose vertices are
$(0,\pm a)$. Notice from the graph that $b = 2$ and the ellipse pass through points $(-1,2)$ which means that the point is a solution on the ellipse.
So, by substitution

$\begin{equation} \begin{aligned} \frac{(-1)^2}{2^2} + \frac{2^2}{a^2} &= 1 \\ \\ \frac{1}{4} + \frac{4}{a^2} &= 1 && \text{Subtract } \frac{1}{4}\\ \\ \frac{4}{a^2} &= \frac{3}{4} && \text{Apply cross multiplication}\\ \\ 3a^2 &= 16 && \text{Solve for } a\\ \\ a &= \frac{4}{\sqrt{3}} \end{aligned} \end{equation}$


Thus, the equation is

$\begin{equation} \begin{aligned} \frac{x^2}{2^2} + \frac{y^2}{\left( \frac{4}{\sqrt{3}} \right)^2} &= 1\\ \\ \text{Or} \\ \\ \frac{x^2}{4} + \frac{y^2}{\frac{16}{3}} &= 1 \\ \\ \frac{x^2}{4} + \frac{3y^2}{16} &= 1 \end{aligned} \end{equation}$