Chapter 10, Problem 31SGA

To determine

To write: The equation of the hyperbola that meets the given set of conditions.

Answer to Problem 31SGA

The equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{20}=1$ .

Explanation of Solution

Given:

The focus is at $\left(-4,-3\right)$ .

The vertices are at $\left(-2,-3\right)$ and $\left(6,-3\right)$ .

Calculation:

From the midpoint of the points $\left(-2,-3\right)$ and $\left(6,-3\right)$ , find the center $\left(h,k\right)$ .

  $\begin{array}{c}\left(h,k\right)=\left(\frac{-2+6}{2},\frac{-3-3}{2}\right)\\ =\left(\frac{4}{2},\frac{-6}{2}\right)\\ =\left(2,-3\right)\end{array}$

From the given coordinates of the vertices, it is known that the transverse axis of the hyperbola is parallel to $x$ axis. So, the vertex formula will be $\left(h\pm a,k\right)$ .

Consider the point $\left(6,-3\right)$ .

  $\begin{array}{c}\left(6,-3\right)=\left(h+a,k\right)\\ 6=h+a\\ 6=2+a\\ a=4\end{array}$

The focus formula will be $\left(h\pm c,k\right)$ .

  $\begin{array}{c}\left(-4,-3\right)=\left(h-c,k\right)\\ -4=h-c\\ -4=2-c\\ c=6\end{array}$

Use the formula $b^2=c^2-a^2$ to find the value of $b$ .

  $\begin{array}{c}{b}^2=6^2-4^2\\ =36-16\\ =20\end{array}$

Find the equation of the hyperbola.

Substitute $2$ for $h$ , $-3$ for $k$ , $16$ for $a^2$ , and $20$ for $b^2$ in the standard equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\frac{{\left(x-2\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{20}=1$

Therefore, the equation of the hyperbola is $\frac{{\left(x-2\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{20}=1$ .