Chapter 13 - Conic Sections - 13.3 Conic Sections: Hyperbolas - 13.3 Exercise Set - Page 871: 7

The graph is shown below

$\frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{16}=1$ Identify a and b by comparing $\frac{{{y}^{2}}}{{{4}^{2}}}-\frac{{{x}^{2}}}{{{4}^{2}}}=1$ with standard equation of hyperbola $\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}}=1$ which has vertical axes. Here, $a=4$ and $b=4$ The asymptotes are, $\begin{align} & y=\frac{b}{a}x \\ & =\frac{4}{4}x \\ & =x \end{align}$ And, $\begin{align} & y=-\frac{b}{a}x \\ & =-\frac{4}{4}x \\ & =-x \end{align}$ When, $a=4$ and $b=4$ then, the possible pairs are as shown below, $\left( -4,4 \right),\left( 4,4 \right),\left( 4,-4 \right),\left( -4,-4 \right)$ 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, $\begin{matrix} x & y=\frac{3}{4}x & y=-\frac{3}{4}x \\ -4 & -3 & 3 \\ 0 & 0 & 0 \\ 1 & \frac{3}{4} & -\frac{3}{4} \\ 4 & 3 & -3 \\ \end{matrix}$ Plot both the asymptotes and square as shown In figure 1 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 $\left( 0,-4 \right)$ and $\left( 0,4 \right)$. Consider the hyperbola equation, $\frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{16}=1$ Rewrite the equation as shown below, $\begin{align} & \frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{16}=1 \\ & \frac{{{y}^{2}}}{16}=1+\frac{{{x}^{2}}}{16} \\ & \frac{{{y}^{2}}}{16}=\frac{16+{{x}^{2}}}{16} \\ & y=\pm \sqrt{16+{{x}^{2}}} \end{align}$ Substitute $x=-4$ into the equation $y=\pm \sqrt{16+{{x}^{2}}}$, $\begin{align} & y=\pm \sqrt{16+{{\left( -4 \right)}^{2}}} \\ & =\pm \frac{3}{4}\sqrt{16+16} \\ & =\pm \frac{3}{4}\left( 5.6568 \right) \\ & =\pm 4.24 \end{align}$ Substitute $x=0$ into the equation $y=\pm \sqrt{16+{{x}^{2}}}$, $\begin{align} & y=\pm \sqrt{16+{{x}^{2}}} \\ & =\pm \sqrt{16} \\ & =\pm \left( 4 \right) \\ & =\pm 4 \end{align}$ Substitute $x=4$ into the equation $y=\pm \sqrt{16+{{x}^{2}}}$, $\begin{align} & y=\pm \sqrt{16+{{\left( 4 \right)}^{2}}} \\ & =\pm \frac{3}{4}\sqrt{16+16} \\ & =\pm \frac{3}{4}\left( 5.6568 \right) \\ & =\pm 4.24 \end{align}$ $\begin{matrix} x & y=\pm \sqrt{16+{{x}^{2}}} \\ -4 & \pm 5.657 \\ 0 & \pm 4 \\ 4 & \pm 5.657 \\ \end{matrix}$ Now plot the graph of the hyperbola as shown in Figure 2.