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Chapter 7: Problem 37

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(y+2)^{2}}{4}-\frac{(x-1)^{2}}{16}=1$$

Short Answer

Expert verified
The center of the hyperbola is at (1, -2), the vertices are at (1, 0) and (1, -4), the foci are at approximately (1, 2.472) and (1, -6.472), and the equations of the asymptotes are \(y = 0.5x - 1.5\) and \(y = -0.5x - 2.5\).

Step by step solution

01

Identify the center

Identify the center (h, k) from the given equation. Here, h = 1 and k = -2 hence the center of the hyperbola is (1, -2).
02

Identify the vertices and the lengths of semi-major and semi-minor axes

The value under (y+2)^2 is a^2; hence a = \sqrt{4} = 2. The value under (x-1)^2 is b^2, hence b = \sqrt{16} = 4. The vertices are situated at a distance of 'a' from the center along the vertical axis. Since the center is (1, -2), the vertices are at (1, -2+2) = (1, 0) and (1, -2-2) = (1, -4).
03

Find the foci

The distance from the center to the foci is given by c = \sqrt{a^2 + b^2}. So, c = \sqrt{4+16} = \sqrt{20} 鈮� 4.472. The foci are at a distance of c along the vertical axis since this is a vertical hyperbola (y-term is first). They would be at (1, -2+4.472) and (1, -2-4.472).
04

Find the equations of the asymptotes

The equations of the asymptotes are given by y = k 卤 (a/b)(x-h). Thus, the equations of the asymptotes are: y = -2 卤 (2/4)(x-1), simplifying to y = -2 卤 0.5x + 0.5, which further simplifies to y = 0.5x - 1.5 and y = -0.5x -2.5.

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