Question

graph the hyperbola.
$9x^2 - 4y^2 = 144$

Explanation:

Step1: Write the equation in standard form

Divide both sides of \(9x^{2}-4y^{2}=144\) by \(144\).
$$\frac{9x^{2}}{144}-\frac{4y^{2}}{144}=\frac{144}{144}$$
Simplify to get \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=1\).

Step2: Identify the parameters

For a hyperbola of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\), here \(a^{2}=16\), so \(a = 4\), and \(b^{2}=36\), so \(b=6\).
The vertices are at \((\pm a,0)\), i.e., \((\pm4,0)\).
The equations of the asymptotes are \(y=\pm\frac{b}{a}x\). Substituting \(a = 4\) and \(b = 6\), we get \(y=\pm\frac{3}{2}x\).

Step3: Graph the hyperbola

• Plot the vertices \((4,0)\) and \((- 4,0)\).
• Draw the asymptotes \(y=\frac{3}{2}x\) (passing through \((0,0)\) with slope \(\frac{3}{2}\)) and \(y =-\frac{3}{2}x\) (passing through \((0,0)\) with slope \(-\frac{3}{2}\)).
• Sketch the hyperbola approaching the asymptotes.

Answer:

The hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{36}=1\) has vertices at \((\pm4,0)\) and asymptotes \(y=\pm\frac{3}{2}x\). Graph by plotting vertices, drawing asymptotes, and sketching the hyperbola approaching the asymptotes.