Chapter 13 - Conic Sections - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 8

The graph is shown below

$\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{9}=1$ Identify a and b by comparing $\frac{{{x}^{2}}}{{{3}^{2}}}-\frac{{{y}^{2}}}{{{3}^{2}}}=1$ with the standard equation of a hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$. Here, $a=3$ and $b=3$ The asymptotes are, $\begin{align} & y=\frac{b}{a}x \\ & =\frac{3}{3}x \\ & =x \end{align}$ And, $\begin{align} & y=-\frac{b}{a}x \\ & =-\frac{3}{3}x \\ & =-x \end{align}$ For $a=3$ and $b=3$, the possible pairs are as shown below, $\left( -3,3 \right),\left( 3,3 \right),\left( 3,-3 \right),\left( -3,-3 \right)$ These pairs form a square, and both asymptotes pass through the corner points of the square. The required table to plot both asymptotes is shown below, $\begin{matrix} x & y=x & y=-x \\ -3 & -3 & 3 \\ 0 & 0 & 0 \\ 3 & 3 & -3 \\ \end{matrix}$ The vertices of the horizontal hyperbola are the mid-points of the right and left sides of the square; thus, the vertices of a horizontal hyperbola are $\left( -3,0 \right)$ and $\left( 3,0 \right)$. Consider the hyperbola equation, $\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{9}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{9}=1 \\ & \frac{{{y}^{2}}}{9}=\frac{{{x}^{2}}}{9}-1 \\ & \frac{{{y}^{2}}}{9}=\frac{{{x}^{2}}-9}{9} \\ & y=\pm \sqrt{{{x}^{2}}-9} \end{align}$ Substitute $x=-4$ into the equation $y=\pm \sqrt{{{x}^{2}}-9}$, $\begin{align} & y=\pm \sqrt{{{\left( -4 \right)}^{2}}-9} \\ & =\pm \sqrt{16-9} \\ & =\pm \sqrt{7} \\ & =\pm 2.646 \end{align}$ Substitute $x=-3$ into the equation $y=\pm \sqrt{{{x}^{2}}-9}$, $\begin{align} & y=\pm \sqrt{{{\left( -3 \right)}^{2}}-9} \\ & =\pm \sqrt{9-9} \\ & =\pm \sqrt{0} \\ & =\pm 0 \end{align}$ Substitute $x=3$ into the equation $y=\pm \sqrt{{{x}^{2}}-9}$, $\begin{align} & y=\pm \sqrt{{{\left( 3 \right)}^{2}}-9} \\ & =\pm \sqrt{9-9} \\ & =\pm \sqrt{0} \\ & =\pm 0 \end{align}$ Substitute $x=4$ into the equation $y=\pm \sqrt{{{x}^{2}}-9}$, $\begin{align} & y=\pm \sqrt{{{\left( 4 \right)}^{2}}-9} \\ & =\pm \sqrt{16-9} \\ & =\pm \sqrt{7} \\ & =\pm 2.646 \end{align}$ $\begin{matrix} x & y=\pm \sqrt{{{x}^{2}}-9} \\ -4 & \pm 2.646 \\ -3 & 0 \\ 3 & 0 \\ 4 & \pm 2.646 \\ \end{matrix}$ Now, plot the graph of the hyperbola as shown in figure 2.