College Algebra, Chapter 8, 8.4, Section 8.4, Problem 22

Find an equation for the conic whose graph is shown.


The hyperbola $\displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ has center on $(h, k)$ and horizontal transverse axis. Based from the graph, the hyperbola has center on $(4, 0)$ since the vertex is equally distant from the center by $2$ units, this gives us $a = 2$. Also, we know that either of the points $(0, 4)$ and $(0, -4)$ satisfy the equation because the hyperbola passes through these points. Solving for $b$, we have


$\begin{equation} \begin{aligned} \frac{(x - 4)^2}{2^2} - \frac{(y - 0)^2}{b^2} =& 1 \\ \\ \frac{(x - 4)^2}{4} - \frac{y^2}{b^2} =& 1 \\ \\ \frac{(x - 4)^2}{4} =& 1 + \frac{y^2}{b^2} \\ \\ \frac{(x - 4)^2}{4} =& \frac{b^2 + y^2}{b^2} \\ \\ b^2 (x - 4)^2 =& 4 (b^2 + y^2) \end{aligned} \end{equation}$


By substituting the point $(0, 4)$,


$\begin{equation} \begin{aligned} b^2 (0 - 4)^2 =& 4(b^2 + 4^2) \\ \\ 16b^2 =& 4b^2 + 64 \\ \\ 12b^2 =& 64 \\ \\ b^2 =& \frac{16}{3} \\ \\ b =& \frac{4}{\sqrt{3}} \end{aligned} \end{equation}$


Therefore, the equation is..

$\displaystyle \frac{(x - 4)^2}{4} - \frac{3y^2}{16} = 1$