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Chapter 7: Problem 42

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$(y-2)^{2}-(x+3)^{2}=5$$

Short Answer

Expert verified
The center of the hyperbola is (-3,2), vertices are located at (-3,2卤sqrt(5)), foci at (-3, 2卤sqrt(10)), and asymptotes are given by the lines y = 卤x + 2.

Step by step solution

01

Identify the Center

The center of the hyperbola is at the point (-3, 2). The minus sign in front of (x+3) makes the x-coordinate of the center negative, while the y-coordinate is the number subtracted from y, which is positive 2.
02

Identify the Vertices and Foci

Because the term with a y is listed first, this hyperbola opens up and down with vertices at points (-3,2卤sqrt(5)). For the foci, the general formula is \((h, k卤sqrt(a^2+b^2))\). Here, because \(a^2=5\) and \(b^2=5\), the foci are (-3, 2卤sqrt(10))
03

Identify the Asymptotes

The equation for asymptotes of a hyperbola that opens up and down is \(y=卤(b/a)(x-h) + k\). Substituting \(a^2=b^2=5\), the equations of asymptotes are y = 卤x + 2.
04

Graphing the Hyperbola

First, plot the center (-3,2), then graph the vertices (-3,2卤sqrt(5)). Sketch the asymptotes with the slope 卤1 and draw the hyperbola opening up and down from the vertices passing through the foci (-3, 2卤sqrt(10)) and approaching the asymptotes.

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