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Chapter 6: Problem 22

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{x^{2}}{36}-\frac{y^{2}}{4}=1$$

Short Answer

Expert verified
The center of the hyperbola is at (0,0), vertices are at (-6, 0) and (6,0), foci are at (卤2鈭�10, 0), and the equations of the asymptotes are y = 卤\(\frac{1}{3}\)x.

Step by step solution

01

Identify the Center

The general form of the horizontal hyperbola is \(\frac{(x鈭抙)^2}{a^2} - \frac{(y鈭択)^2}{b^2}=1\). The center of the hyperbola is (h, k). From the given equation one can see that the center is at (0, 0) since there is no subtraction in the binomials.
02

Identify the Vertices

The vertices are (卤a, 0), using the value of a found from the equation which is the square root of the denominator under x. Hence, a = 鈭�36 = 6, the vertices are (卤6, 0) or (-6, 0) and (6,0).
03

Identify the Foci

The foci are located at (卤c, 0). c can be found using the formula \(c=\sqrt{a^{2}+b^{2}}\). In this equation, a = 6 and b = 2. On substituting the values, we find the values of c to be 卤\(\sqrt{36 + 4}\) = 卤\(\sqrt{40}\) = 卤2鈭�10. Therefore, the foci are at (卤2鈭�10, 0).
04

Find the equations of the asymptotes

The asymptotes of the hyperbola are the lines y = 卤\(\frac{b}{a}\)x . Here, a = 6 and b = 2. Hence, the equations of asymptotes are y = 卤\(\frac{2}{6}\)x = 卤\(\frac{1}{3}\)x.
05

Sketch the Hyperbola

Sketch the hyperbola by marking the center, vertices, and foci, and drawing the asymptotes. The hyperbola opens to the right and left because the x虏 term has a positive coefficient. Plot the vertices and foci on the x-axis, and draw the asymptote lines y = 卤\(\frac{1}{3}\)x. Draw smooth curves extending from the vertices to the asymptotes to complete the sketch.

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