Chapter 9.3, Problem 12E

To determine

To find: The standard form of the equation of the hyperbola when vertices are $\left(0,\pm 2\right)$ and foci is $\left(0,\pm 4\right)$ .

Answer to Problem 12E

The standard form of the equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{27}=1$ .

Explanation of Solution

Given information:

The given characteristics is Vertices: $\left(0,\pm 2\right)$ ; foci: $\left(0,\pm 4\right)$ .

Calculation:

The center is the origin. The distance between one of the foci and the center is $c=6$ .

The vertices lie on the horizontal line, so the hyperbola has a horizontal transverse axis. The distance between one of the vertices and the center is $a=3$ .

Substitute $3$ for $a$ and $6$ for $c$ in the equation $b=\sqrt{c^2-a^2}$ to find $b$ .

  $\begin{array}{c}b=\sqrt{6^2-3^2}\\ =\sqrt{36-9}\\ =\sqrt{27}\end{array}$

Substitute $3$ for $a$ , $\sqrt{27}$ for $b$ , $0$ for $h$ and $0$ for $k$ in the equation of a hyperbola with a horizontal transverse axis $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\begin{array}{c}\frac{{\left(x-0\right)}^2}{3^2}-\frac{{\left(y-0\right)}^2}{{\left(\sqrt{27}\right)}^2}=1\\ \frac{x^2}{9}-\frac{y^2}{27}=1\end{array}$

Therefore, the standard form of the equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{27}=1$ .