Chapter 8.5, Problem 13E

Solution

Graph the given equation and identify the vertices, foci, and asymptote of the hyperbola.

The graph of hyperbola is

  

with vertices $\left(\pm 4,0\right)$ , foci $\left(0,\pm \sqrt{97}\right)$ , and asymptote at $x=\pm \frac{4}{9}y$ .

Given information:

Given equation is:

  $81x^2-16y^2=1296$

Formula used:

• Standard form of equation of a hyperbola with horizontal major axis and vertices $\left(\pm a,0\right)$ and co- vertices $\left(0,\pm b\right)$ is

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

And, the asymptote of the hyperbola is calculated as

  $x=\pm \frac{a}{b}y$

• Standard form of equation of a hyperbola with vertical major axis and vertices as $\left(0,\pm a\right)$ and co- vertices $\left(\pm b,0\right)$ is

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

And, the asymptote of the hyperbola is calculated as

  $y=\pm \frac{a}{b}x$

• Foci of hyperbola with vertices as $\left(\pm a,0\right)$ or $\left(0,\pm a\right)$ and co- vertices $\left(0,\pm b\right)$ or $\left(\pm b,0\right)$ is found as:

  $c^2=a^2+b^2$

Calculation:

The given equation is

  $81x^2-16y^2=1296$

First divide the above equation on both sides by 1296 to make it in the standard form.

  $\begin{array}{l}\frac{81x^2-16y^2}{1296}=\frac{1296}{1296}\\ \Rightarrow \frac{81x^2}{1296}-\frac{16y^2}{1296}=1\\ \Rightarrow \frac{x^2}{16}-\frac{y^2}{81}=1\end{array}$

Now in order to graph the given equation of hyperbola

  $\frac{x^2}{16}-\frac{y^2}{81}=1$ ,

Simplify it, as

  $\frac{x^2}{4^2}-\frac{y^2}{9^2}=1$

Now, identify its vertices, foci, and asymptote.

Since here $x^2$ term is positive, i.e. comparing the given equation with

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

gives

  $a=4,b=9$ , and the transverse axis is horizontal with vertices $\left(\pm 4,0\right)$ , and co-vertices is

  $\left(0,\pm 9\right)$ .

Now, find the foci using the formula

  $\begin{array}{c}{c}^2=a^2+b^2\\ c^2=4^2+9^2\\ c^2=16+81\\ c=\pm \sqrt{97}\end{array}$

Thus, foci are at $\left(0,\pm \sqrt{97}\right)$

And, the asymptote of the hyperbola is calculated as

  $\begin{array}{c}x=\pm \frac{a}{b}y\\ =\pm \frac{4}{9}y\end{array}$

Now, draw the graph hyperbola by drawing the rectangle centered at origin with width $2a=8$ and $2b=18$ and draw asymptotes that passes through corner of this rectangle. And draw the hyperbola passing through the vertices and approaching the asymptotes.

Thus, the graph of the hyperbola will be as

  

with vertices

  $\left(\pm 4,0\right)$ , foci $\left(0,\pm \sqrt{97}\right)$ , and asymptote at $x=\pm \frac{4}{9}y$ .