Chapter 9.3, Problem 19E

To determine

To find: The standard form of the equation of the hyperbola whenvertices are $\left(0,4\right),\left(0,0\right)$ and passes through the point $\left(\sqrt{5},-1\right)$ .

Answer to Problem 19E

The standard form of the equation of the hyperbola is $\frac{{\left(y-2\right)}^2}{4}-\frac{x^2}{4}=1$ .

Explanation of Solution

Given information:

The given characteristics is Vertices: $\left(0,4\right),\left(0,0\right)$ passes through the point $\left(\sqrt{5},-1\right)$ .

Calculation:

The center is at the midpoint of the two vertices. The coordinate of the center is $\left(0,2\right)$ .

The distance between one of the vertices and the center is $a=2$ .

The vertices lie on a vertical line, so the hyperbola has a vertical transverse axis. Use the formula for a hyperbola with a vertical transverse axis.

  $\frac{{\left(y-2\right)}^2}{4}-\frac{x^2}{b^2}=1$

Substitute $-1$ for $y$ , and $\sqrt{5}$ for $x$ in the above equation to find $b$ .

  $\begin{array}{c}\frac{{\left(-1-2\right)}^2}{4}-\frac{{\left(\sqrt{5}\right)}^2}{b^2}=1\\ \frac{9}{4}-\frac{5}{b^2}=1\\ b^2=4\\ b=2\end{array}$

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

  $\begin{array}{c}\frac{{\left(y-2\right)}^2}{2^2}-\frac{{\left(x-0\right)}^2}{2^2}=1\\ \frac{{\left(y-2\right)}^2}{4}-\frac{x^2}{4}=1\end{array}$

Therefore, the standard form of the equation of the hyperbola is $\frac{{\left(y-2\right)}^2}{4}-\frac{x^2}{4}=1$ .