Chapter 13 - Conic Sections - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 12

The graph is shown below

$\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{36}=1$ Identify a and b by comparing $\frac{{{x}^{2}}}{{{5}^{2}}}-\frac{{{y}^{2}}}{{{6}^{2}}}=1$ with the standard equation $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$ which has horizontal axes. Here, $a=5$ and $b=6$ The asymptotes are, $\begin{align} & y=\frac{b}{a}x \\ & =\frac{6}{5}x \end{align}$ And, $\begin{align} & y=-\frac{b}{a}x \\ & =-\frac{6}{5}x \end{align}$ $\left( -5,6 \right),\left( 5,6 \right),\left( 5,-6 \right),\left( -5,-6 \right)$ 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, $\begin{matrix} x & y=\frac{6}{5}x & y=-\frac{6}{5}x \\ -5 & -6 & 6 \\ 0 & 0 & 0 \\ 5 & 6 & -6 \\ \end{matrix}$ Plot the asymptotes and rectangle as shown below in Figure - 1. The vertices of the horizontal hyperbola are the mid-points of the left and right sides of the rectangle; thus, the vertices of the horizontal hyperbola are $\left( -5,0 \right)$ and $\left( 5,0 \right)$. Consider the hyperbola equation, $\frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{36}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{x}^{2}}}{25}-\frac{{{y}^{2}}}{36}=1 \\ & \frac{{{y}^{2}}}{36}=\frac{{{x}^{2}}}{25}-1 \\ & \frac{{{y}^{2}}}{36}=\frac{{{x}^{2}}-25}{25} \\ & y=\pm \frac{6}{5}\sqrt{{{x}^{2}}-25} \end{align}$ Substitute $x=-10$ into the equation $y=\pm \frac{6}{5}\sqrt{{{x}^{2}}-25}$, $\begin{align} & y=\pm \frac{6}{5}\sqrt{{{\left( -10 \right)}^{2}}-25} \\ & =\pm \frac{6}{5}\sqrt{100-25} \\ & =\pm \frac{6}{5}\left( 8.66 \right) \\ & =\pm 10.392 \end{align}$ Substitute $x=-5$ into the equation $y=\pm \frac{6}{5}\sqrt{{{x}^{2}}-25}$, $\begin{align} & y=\pm \frac{6}{5}\sqrt{{{\left( -5 \right)}^{2}}-25} \\ & =\pm \frac{6}{5}\sqrt{25-25} \\ & =\pm \frac{6}{5}\left( 0 \right) \\ & =0 \end{align}$ Substitute $x=5$ into the equation $y=\pm \frac{2}{5}\sqrt{{{x}^{2}}-25}$, $\begin{align} & y=\pm \frac{6}{5}\sqrt{{{\left( 5 \right)}^{2}}-25} \\ & =\pm \frac{6}{5}\sqrt{25-25} \\ & =\pm \frac{6}{5}\left( 0 \right) \\ & =0 \end{align}$ Substitute $x=10$ into the equation $y=\pm \frac{2}{5}\sqrt{{{x}^{2}}-25}$, $\begin{align} & y=\pm \frac{6}{5}\sqrt{{{\left( 10 \right)}^{2}}-25} \\ & =\pm \frac{6}{5}\sqrt{100-25} \\ & =\pm \frac{6}{5}\left( 8.66 \right) \\ & =\pm 10.392 \end{align}$ The required table is: $\begin{matrix} x & y=\pm \frac{2}{5}\sqrt{{{x}^{2}}-25} \\ -10 & \pm 10.392 \\ -5 & 0 \\ 5 & 0 \\ 10 & \pm 10.392 \\ \end{matrix}$ Now plot the hyperbola.