Chapter 10.4, Problem 21E

To determine

To find:the center, vertices, foci, and the equation of the asymptotes of the hyperbola. And sketch hyperbola using asymptotes.

Answer to Problem 21E

Thehyperbola has centerat $\left(0,0\right)$ , verticesat $\left(0,\pm 5\right)$ , foci $\left(0,\pm \sqrt{106}\right)$ , and the equation of the asymptotes is $y=\pm \frac{5}{9}x$ .

Explanation of Solution

Given information:

The given equation is $\frac{x^2}{25}-\frac{y^2}{81}=1$ .

Calculation:

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

  $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$.......(1)

Rearrange the equation (1).

  $\frac{{\left(y-0\right)}^2}{{\left(5\right)}^2}-\frac{{\left(x-0\right)}^2}{{\left(9\right)}^2}=1$.......(2)

Compare the equation (1) and (2).

  $\begin{array}{l}a=5\\ b=9\end{array}$

The center is at $\left(0,0\right)$ .

Calculate the value of $c$ .

  $\begin{array}{c}c=\sqrt{a^2+b^2}\\ =\sqrt{5^2+9^2}\\ =\sqrt{106}\end{array}$

The vertices are $\left(h,\pm a+k\right)$ , foci at $\left(h,\pm c+k\right)$ .

Then the vertices will be $\left(0,\pm 5\right)$ .

And the foci will be $\left(0,\pm \sqrt{106}\right)$ .

Calculate the equation of the asymptotes of the hyperbola.

  $\begin{array}{l}y=k\pm \frac{a}{b}\left(x-h\right)\\ y=\pm \frac{5}{9}x\end{array}$ .

The graph for the asymptotes is shown in figure (1).

  

Figure (1)

The graph of $\frac{x^2}{25}-\frac{y^2}{81}=1$ using asymptotes is shown in figure (2).

  

Figure (2)

Therefore, the hyperbola has centerat $\left(0,0\right)$ , verticesat $\left(0,\pm 5\right)$ , foci $\left(0,\pm \sqrt{106}\right)$ , and the equation of the asymptotes is $y=\pm \frac{5}{9}x$ .