Problem 42 - College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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±sqrt(5)). For the foci, the general formula is \((h, k±sqrt(a^2+b^2))\). Here, because \(a^2=5\) and \(b^2=5\), the foci are (-3, 2±sqrt(10))

Step 3: Identify the Asymptotes

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

Step 4: Graphing the Hyperbola

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