Problem 38 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1$$

Step-by-Step Solution

Answer

The center of the hyperbola is at (-1, 2), the vertices are at (-1, -4) and (-1, 8), and the asymptotes are at y = ±(6/7)x. The foci are at (-1, 2±√85).

Step 1: Identify the Center

The center of the hyperbola can be identified from the numbers that are added or subtracted from y and x in the equation \(\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1\). Thus, the center of this hyperbola is: (-1, 2)

Step 2: Find the Vertices and Asymptotes

The vertices and asymptotes are determined from the denominators of the y and x terms in the equation. The square root of these numbers gives the distances from the center to the vertices along the y-axis, and the distance from the center to the asymptotes along the x-axis. Thus, square root of 36 yields 6, and square root of 49 yields 7. Thus, the vertices are at (-1, 2±6) = (-1, -4) and (-1, 8) and the asymptotes are at y = ±(6/7)x.

Step 3: Locate the Foci

The foci of a hyperbola are points lying on the principal axis of the hyperbola and are found using the equation \(c = \sqrt{a^2 + b^2} \), where a is the distance to the vertices (6) and b is the distance to the asymptotes (7). Thus, substituting the values, \(c = \sqrt{6^2 + 7^2} = \sqrt{85}\). Thus, the foci are at (-1, 2±√85).

Step 4: Graph the Hyperbola

Now, the hyperbola can be graphed using the center, vertices, asymptotes, and foci, on the coordinate plane.