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Chapter 6: Problem 30

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$x^{2}-9 y^{2}+36 y-72=0$$

Short Answer

Expert verified
The center of the hyperbola is (0,2), vertices are (卤8.48,0), foci are (卤8.94,0), and equations of the asymptotes are \( y = 2 \pm \frac{鈭�2x}{3} \).

Step by step solution

01

Rewrite Equation in Standard Form

First, rewrite the equation given to standard form. Begin by isolating the \( y^{2} \) term:\( x^{2}-9(y^2-4y)-72=0\)Add 36, which is the square of half of the coefficient of \( y \), to both sides of the equation to complete the square:\( x^{2}-9(y-2)^2=72 \)Finally, divide both sides by 72 to get the equation in standard form:\( \frac{x^{2}}{72} - \frac{(y-2)^2}{8} = 1\)
02

Calculate Key Elements of the Hyperbola

From the standard form, identify \( h = 0, k = 2, a^2 = 72, b^2 = 8 \). Then calculate \( c \) using \( c^2 = a^2 + b^2 \) to get \( c = \sqrt{80} \).The center is (h, k) = (0,2). The vertices are (a,0) = (卤鈭�72,0), i.e., (卤8.48,0).The foci are (c,0) = (卤鈭�80,0), i.e., (卤8.94,0).The equations of the asymptotes are \( y = k \pm \frac{b}{a}(x-h) = 2 \pm \frac{鈭�8}{鈭�72}x \) or \( y = 2 \pm \frac{鈭�2x}{3} \).
03

Sketch the Hyperbola

Start with the center at (0,2). Plot the vertices (卤8.48,0) and the foci (卤8.94,0). The hyperbola opens horizontally since \( x \) is positive in the equation. Draw the asymptotes with the slopes of 卤鈭�2/3 to help shape the hyperbola. The hyperbola should touch the vertices and approach the asymptotes as \( x \) moves towards 卤鈭�.

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