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Chapter 6: Problem 37

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Use a graphing utility to graph the hyperbola and its asymptotes. $$9 y^{2}-x^{2}+2 x+54 y+62=0$$

Short Answer

Expert verified
The center of the hyperbola is (1,-3), the vertices are at distances 3 units above and below the center, the foci are at \(\sqrt{10}\) units below and above the center, and the asymptotes are y = -3 + 3(x - 1) and y = -3 - 3(x - 1). Use a graphing utility to visualize the hyperbola and its asymptotes.

Step by step solution

01

Rearrange the equation in standard form

The equation of a hyperbola in standard form is \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2} = 1\) for a hyperbola that opens left and right or \((y-k)^{2}/a^{2} - (x-h)^{2}/b^{2} = 1\) for a hyperbola that opens up and down, where (h,k) is the center, a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices. So, \[9y^{2}-x^{2}+2x+54y+62=0\] should be rearranged as: \[(y+3)^2-\frac{(x-1)^2}{9}=1\].
02

Find the center, vertices, and foci

By comparing with the standard form, we can identify that the center (h,k) is (1,-3). The value of a (distance from center to vertices) is 3 (which is under the y term) and b (distance from center to co-vertices) is 1 (which is under the x term with a square root). The foci can be calculated using the formula \(c = \sqrt{a^{2} + b^{2}}\). Hence c = \(\sqrt{3^2 + 1^2}\) = \(\sqrt{10}\).
03

Calculate the equations of the asymptotes

The equations of the asymptotes for a hyperbola that opens up and down are given by \(y = k \pm (a/b)(x - h)\). So the equations of the asymptotes are y = -3 \(\pm\) 3(x - 1).
04

Graph the hyperbola and its asymptotes

Use a graphing software to plot and visualize the hyperbola by plotting the center, vertices, foci, and the asymptotes calculated in the previous steps.

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