Chapter 9.3, Problem 49E

To determine

To find: The standard form of the equation of hyperbola with the given characteristics.

Answer to Problem 49E

The standard form of the equation of the hyperbola is $\frac{{\left(x-4\right)}^2}{16}-\frac{{\left(y-3\right)}^2}{9}=1$ .

Explanation of Solution

Given information:

The characteristics are foci: $\left(-1,3\right),\left(9,3\right)$ ; asymptotes: $y=\frac{3}{4}x,y=6-\frac{3}{4}x$ .

Calculation:

For a right-left facing hyperbola, the foci are defined as $\left(h+c,k\right)$ and $\left(h-c,k\right)$ .

Solve for center $\left(h,k\right)$ by comparing with the coordinates of given foci.

  $h+c=-1$.........(1)

  $h-c=9$.........(2)

  $k=3$

Add equation (1) and (2) to find $h$ .

  $\begin{array}{c}\left(h+c\right)+\left(h-c\right)=-1+9\\ 2h=8\\ h=\frac{8}{2}\\ =4\end{array}$

The center of the hyperbola will be $\left(4,3\right)$ .

Solve for $a$ and b by comparing the equation to the two given asymptotes.

  $\begin{array}{c}a=4\\ b=3\end{array}$

Substitute 4 for $h$ , 3 for $k$ , 4 for $a$ and 3 for $b$ in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ to find the standard form of hyperbola equation.

  $\begin{array}{c}\frac{{\left(x-4\right)}^2}{4^2}-\frac{{\left(y-3\right)}^2}{3^2}=1\\ \frac{{\left(x-4\right)}^2}{16}-\frac{{\left(y-3\right)}^2}{9}=1\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{{\left(x-4\right)}^2}{16}-\frac{{\left(y-3\right)}^2}{9}=1$ .