Chapter 13 - Conic Sections - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 9

The graph is shown below

$\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{25}=1$ 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, $a=2$ and $b=5$ The asymptotes are, $\begin{align} & y=\frac{b}{a}x \\ & =\frac{5}{2}x \end{align}$ And, $\begin{align} & y=-\frac{b}{a}x \\ & =-\frac{5}{2}x \end{align}$ For $a=2$ and $b=5$, the possible pairs are as shown below: $\left( -2,5 \right),\left( 2,5 \right),\left( 2,-5 \right),\left( -2,-5 \right)$ 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, $\begin{matrix} x & y=\frac{5}{2}x & y=-\frac{5}{2}x \\ -2 & -5 & 5 \\ 0 & 0 & 0 \\ 2 & 5 & -5 \\ \end{matrix}$ 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 $\left( -2,0 \right)$ and $\left( 2,0 \right)$. Consider the hyperbola equation, $\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{25}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{25}=1 \\ & \frac{{{y}^{2}}}{25}=\frac{{{x}^{2}}}{4}-1 \\ & \frac{{{y}^{2}}}{4}=\frac{{{x}^{2}}-4}{4} \\ & y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4} \end{align}$ Substitute $x=-5$ into the equation $y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4}$, $\begin{align} & y=\pm \frac{5}{2}\sqrt{{{\left( -5 \right)}^{2}}-4} \\ & =\pm \frac{5}{2}\sqrt{25-4} \\ & =\pm \frac{5}{2}\left( 4.58 \right) \\ & =\pm 11.46 \end{align}$ Substitute $x=-2$ into the equation $y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4}$, $\begin{align} & y=\pm \frac{5}{2}\sqrt{{{\left( -2 \right)}^{2}}-4} \\ & =\pm \frac{5}{2}\sqrt{4-4} \\ & =\pm \frac{5}{2}\left( 0 \right) \\ & =0 \end{align}$ Substitute $x=2$ into the equation $y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4}$, $\begin{align} & y=\pm \frac{5}{2}\sqrt{{{\left( 2 \right)}^{2}}-4} \\ & =\pm \frac{5}{2}\sqrt{4-4} \\ & =\pm \frac{5}{2}\left( 0 \right) \\ & =0 \end{align}$ Substitute $x=5$ into the equation $y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4}$, $\begin{align} & y=\pm \frac{5}{2}\sqrt{{{\left( 5 \right)}^{2}}-4} \\ & =\pm \frac{5}{2}\sqrt{25-4} \\ & =\pm \frac{5}{2}\left( 4.58 \right) \\ & =\pm 11.46 \end{align}$ $\begin{matrix} x & y=\pm \frac{5}{2}\sqrt{{{x}^{2}}-4} \\ -5 & \pm 11.46 \\ -2 & 0 \\ 2 & 0 \\ 5 & \pm 11.46 \\ \end{matrix}$ Now plot the graph of the hyperbola as shown in Figure - 2.