Chapter 13 - Conic Sections - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 11

The graph is shown below.

$\frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1$ 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 $b=6$ The asymptotes are, $\begin{align} & y=\frac{b}{a}x \\ & =\frac{6}{3}x \\ & =2x \end{align}$ And, $\begin{align} & y=-\frac{b}{a}x \\ & =-\frac{6}{3}x \\ & =-2x \end{align}$ For $a=3$ and $b=6$, the possible pairs are as shown below, $\left( -3,6 \right),\left( 3,6 \right),\left( 3,-6 \right),\left( -3,-6 \right)$ These pairs form a rectangle, and both asymptotes pass through the corner points of the rectangle. The required table to plot both asymptotes is shown below, $\begin{matrix} x & y=2x & y=-2x \\ -3 & -6 & 6 \\ 0 & 0 & 0 \\ 3 & 6 & -6 \\ \end{matrix}$ The vertices of the vertical hyperbola are the mid-points of the top and bottom sides of the rectangle; thus, the vertices of the vertical hyperbola are $\left( 0,-6 \right)$ and $\left( 0,6 \right)$. Consider the hyperbola equation, $\frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1 \\ & \frac{{{y}^{2}}}{36}=1+\frac{{{x}^{2}}}{9} \\ & \frac{{{y}^{2}}}{36}=\frac{9+{{x}^{2}}}{9} \\ & y=\pm 2\sqrt{9+{{x}^{2}}} \end{align}$ Substitute $x=-5$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( -5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ Substitute $x=0$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 0 \right)}^{2}}} \\ & =\pm 2\sqrt{9} \\ & =\pm 2\left( 3 \right) \\ & =\pm 6 \end{align}$ Substitute $x=5$ into the equation, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ $\begin{matrix} x & y=\pm \frac{4}{3}\sqrt{9+{{x}^{2}}} \\ -5 & \pm 11.66 \\ 0 & \pm 6 \\ 5 & \pm 11.66 \\ \end{matrix}$ Now plot the graph of the hyperbola as shown below. The vertices of the vertical hyperbola are the mid-points of the top and bottom sides of the rectangle; thus, the vertices of the vertical hyperbola are $\left( 0,-6 \right)$ and $\left( 0,6 \right)$. Consider the hyperbola equation, $\frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1 \\ & \frac{{{y}^{2}}}{36}=1+\frac{{{x}^{2}}}{9} \\ & \frac{{{y}^{2}}}{36}=\frac{9+{{x}^{2}}}{9} \\ & y=\pm 2\sqrt{9+{{x}^{2}}} \end{align}$ Substitute $x=-5$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( -5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ Substitute $x=0$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 0 \right)}^{2}}} \\ & =\pm 2\sqrt{9} \\ & =\pm 2\left( 3 \right) \\ & =\pm 6 \end{align}$ Substitute $x=5$ into the equation, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ The required table is: $\begin{matrix} x & y=\pm \frac{4}{3}\sqrt{9+{{x}^{2}}} \\ -5 & \pm 11.66 \\ 0 & \pm 6 \\ 5 & \pm 11.66 \\ \end{matrix}$ Consider the hyperbola equation, $\frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{y}^{2}}}{36}-\frac{{{x}^{2}}}{9}=1 \\ & \frac{{{y}^{2}}}{36}=1+\frac{{{x}^{2}}}{9} \\ & \frac{{{y}^{2}}}{36}=\frac{9+{{x}^{2}}}{9} \\ & y=\pm 2\sqrt{9+{{x}^{2}}} \end{align}$ Substitute $x=-5$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( -5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ Substitute $x=0$ into the equation $y=\pm 2\sqrt{9+{{x}^{2}}}$, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 0 \right)}^{2}}} \\ & =\pm 2\sqrt{9} \\ & =\pm 2\left( 3 \right) \\ & =\pm 6 \end{align}$ Substitute $x=5$ into the equation, $\begin{align} & y=\pm 2\sqrt{9+{{\left( 5 \right)}^{2}}} \\ & =\pm 2\sqrt{9+25} \\ & =\pm 2\left( 5.83 \right) \\ & =\pm 11.66 \end{align}$ The required table is: $\begin{matrix} x & y=\pm \frac{4}{3}\sqrt{9+{{x}^{2}}} \\ -5 & \pm 11.66 \\ 0 & \pm 6 \\ 5 & \pm 11.66 \\ \end{matrix}$ Now plot the graph.