Example 4 from Algebra and Trigonometry, 12.2 The Hyperbola

Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form

Graph the hyperbola given by the equation $\frac{y^{2}}{64} - \frac{x^{2}}{36} = 1.$ Identify and label the vertices, co-vertices, foci, and asymptotes.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The standard form that applies to the given equation is $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1.$ Thus, the transverse axis is on the y-axis
The coordinates of the vertices are $(0, \pm a) = (0, \pm \sqrt{64}) = (0, \pm 8)$

The coordinates of the co-vertices are $(\pm b, 0) = (\pm \sqrt{36}, 0) = (\pm 6, 0)$

The coordinates of the foci are $(0, \pm c),$ where $c = \pm \sqrt{a^{2} + b^{2}} .$ Solving for $c,$ we have

$c = \pm \sqrt{a^{2} + b^{2}} = \pm \sqrt{64 + 36} = \pm \sqrt{100} = \pm 10$

Therefore, the coordinates of the foci are $(0, \pm 10)$
The equations of the asymptotes are $y = \pm \frac{a}{b} x = \pm \frac{8}{6} x = \pm \frac{4}{3} x$
Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 8.

Figure 8

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Graph the hyperbola given by the equation $\frac{x^{2}}{144} - \frac{y^{2}}{81} = 1.$ Identify and label the vertices, co-vertices, foci, and asymptotes.

How did it go?

Solution

Your turn

Solution

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