Chapter 10.4, Problem 42E

To determine

To write: An equation of the hyperbola whose slopes of asymptotes are $\pm 2$ and the foci are at $\left(1,5\right)\text{ and}\left(1,-3\right)$ .

Answer to Problem 42E

  $\frac{5{\left(y-1\right)}^2}{64}-\frac{5{\left(x-1\right)}^2}{16}=1$

Explanation of Solution

Given information: Slopes of asymptotes are $\pm 2$ and the foci are at $\left(1,5\right)\text{ and}\left(1,-3\right)$

Calculation:

The foci tell us the transverse axis is vertical, the center is at $\left(1,1\right)$ and c = 4.

Slopes of the asymptotes being $\pm 2$ tell us that $\frac{a}{b}=2$ .

Using this information we can make:

  $\begin{array}{l}a=2b\to {\left(2b\right)}^2+b^2=16\\ b^2=\frac{16}{5}\end{array}$

Since,

  $\begin{array}{l}a=2b\\ a=\frac{2\left(4\right)}{\sqrt{5}}=\frac{8}{\sqrt{5}}\end{array}$

This gives us:

  $\begin{array}{l}\frac{{\left(y-1\right)}^2}{\frac{64}{5}}-\frac{{\left(x-1\right)}^2}{\frac{16}{5}}=1\text{ or }\\ \frac{5{\left(y-1\right)}^2}{64}-\frac{5{\left(x-1\right)}^2}{16}=1\end{array}$