Chapter 8.4, Problem 1E

Solution

To calculate: The vertices and foci of the hyperbola $xy=4$ by analyzing the graph.

The vertices of the hyperbola $xy=4$ are $\left(-2,-2\right)$ and $\left(2,2\right)$ , and the foci are $\left(-2\sqrt{2},-2\sqrt{2}\right)$ and $\left(2\sqrt{2},2\sqrt{2}\right)$ .

Given information: The given equation of the hyperbola is $xy=4$ . The graph is symmetric about the lines $y=x$ and $y=-x$ .

Calculation:

Draw the graph of the hyperbola $xy=4$ and lines of symmetry as shown below.

  

Figure (1)

From the graph, it can be observed that the vertices of the hyperbola are $\left(-2,-2\right)$ and $\left(2,2\right)$ .

The foci lie on the line $y=x$ and $c$ units from the origin. The distance of a vertex to the center is $a$ . Consider the vertex $\left(2,2\right)$ .

Calculate the value of $a$ .

  $\begin{array}{c}a=\sqrt{2^2+2^2}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

If the hyperbola is horizontal, its asymptotes are $y=\pm x$ . So, $\frac{b}{a}=1$ .

  $\begin{array}{c}a=b\\ b=2\sqrt{2}\end{array}$

Calculate the value of $c$ by the formula $c^2=a^2+b^2$ .

  $\begin{array}{c}c=\sqrt{{\left(2\sqrt{2}\right)}^2+{\left(2\sqrt{2}\right)}^2}\\ =\sqrt{8+8}\\ =\sqrt{16}\\ =4\end{array}$

So, the foci is at $4$ units distance from the center along the line $y=x$ . Let the foci be $\left(-m,-m\right)$ and $\left(m,m\right)$ .

Apply Pythagorean theorem and find the value of $n$ .

  $\begin{array}{c}{m}^2+m^2=4^2\\ 2m^2=16\\ m^2=8\\ m=2\sqrt{2}\end{array}$

The foci of the hyperbola are $\left(-2\sqrt{2},-2\sqrt{2}\right)$ and $\left(2\sqrt{2},2\sqrt{2}\right)$ .

Thus, the vertices of the hyperbola $xy=4$ are $\left(-2,-2\right)$ and $\left(2,2\right)$ , and the foci are $\left(-2\sqrt{2},-2\sqrt{2}\right)$ and $\left(2\sqrt{2},2\sqrt{2}\right)$ .