Chapter 9.7, Problem 63SR

To determine

To find: The coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation and then graph the hyperbola.

Answer to Problem 63SR

The coordinates of the vertices are $\left(2,8\right)\text{and}\left(2,-2\right)$ , the coordinates of the foci is $\left(2,3\pm \sqrt{41}\right)$ , the equation of asymptote is $y=\pm \frac{5}{4}\left(x-2\right)+3$ , and the graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given:

The equation of the hyperbola is $\frac{{\left(y-3\right)}^2}{25}-\frac{{\left(x-2\right)}^2}{16}=1$ .

Calculation:

The given hyperbola is a vertical hyperbola.

Compare the given equation with general equation $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ to get $\left(h,k\right)=\left(2,3\right)$ .

  $\begin{array}{c}{a}^2=25\\ a=5\\ b^2=16\\ =4\end{array}$

Use the relation between $a$ , $b$ and $c$ to compute the value of $c$ as follows.

  $\begin{array}{c}{c}^2=a^2+b^2\\ =25+16\\ =41\\ c=\sqrt{41}\end{array}$

Find the coordinates of the vertices of the hyperbola as follows.

  $\begin{array}{c}\left(h,k\pm a\right)=\left(2,3\pm 5\right)\\ =\left(2,8\right)\text{and}\left(2,-2\right)\end{array}$

Find the coordinates of the foci of the hyperbola as follows.

  $\begin{array}{c}\left(h,k\pm c\right)=\left(2,3\pm \sqrt{41}\right)\\ =\left(2,3\pm \sqrt{41}\right)\end{array}$

Find the equation of asymptotes as follows.

  $\begin{array}{c}y-k=\pm \frac{a}{b}\left(x-h\right)\\ y-3=\pm \frac{5}{4}\left(x-2\right)\\ y=\pm \frac{5}{4}\left(x-2\right)+3\end{array}$

The graph of the given parabola is as follows.

  

Figure (1)

Therefore, the coordinates of the vertices are $\left(2,8\right)\text{and}\left(2,-2\right)$ , the coordinates of the foci is $\left(2,3\pm \sqrt{41}\right)$ , the equation of asymptote is $y=\pm \frac{5}{4}\left(x-2\right)+3$ , and the graph of the hyperbola is shown in figure (1).