Graphing a Hyperbola With Center at the Origin

Graph the equation \(\dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}=1\).

Solution  The graph of \(\dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}=1\) is a hyperbola. The hyperbola consists of two branches, one opening up, the other opening down, like the graph in Figure 41(b). The hyperbola has no \(x\) -intercepts. To find the \(y\)-intercepts, we let \(x=0\) and solve for \(y.\) \[ \begin{array}{l} \dfrac{y^{2}}{4}=1 \\[11pt] y^{2}=4 \\[5pt] y=-2\qquad\hbox{ or} \qquad y=2 \end{array} \]

A-25

The \(y\)-intercepts are \(-2\) and \(2\), so the vertices are \((0,-2)\) and (\(0,2\)). The transverse axis is the vertical line \(x=0.\) To graph the hyperbola, let \(y=\pm\, 3\) (or any numbers \(\geq\)\(2\) or \(\leq\)\(-2\)). Then \[ \begin{array}{rclll} \dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}&=&1 & & \\ \dfrac{9}{4}-\dfrac{x^{2}}{5}&=&1 & & {\color{#0066A7}{y=\pm\, 3}} \\ ~\ \ \ \ \ \ \dfrac{x^{2}}{5}&=&\dfrac{5}{4} & & \\ ~\ \ \ \ \ \ \ x^{2}&=&\dfrac{25}{4} & & \\ x=-\dfrac{5}{2} &\hbox{or}& x=\dfrac{5}{2} \\ \end{array} \]

The points \(\left( -\dfrac{5}{2},3\right) \), \(\left( -\dfrac{5}{2},-3\right) \), \(\left( \dfrac{5}{2},3\right) \), and \(\left( \dfrac{5}{2},-3\right)\) are on the hyperbola. See Figure 42 for the graph.