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Chapter 6: Problem 28

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{(y-1)^{2}}{1 / 4}-\frac{(x+3)^{2}}{1 / 16}=1$$

Short Answer

Expert verified
The center of the hyperbola is (-3, 1), the vertices are (-3, 1/2) and (-3, 3/2), the foci are (-3, -1/4) and (-3, 9/4), and the equations of the asymptotes are y = 1 + x/2 and y = 1 - x/2.

Step by step solution

01

Identify the Center

Looking at our equation, we can see that h = -3 and k = 1. So the center of the hyperbola is (-3, 1).
02

Find the Vertices

The term with y has the positive sign and a square of 1/4, so the hyperbola is vertically oriented and a = 1/2. The vertices are therefore (h, k 卤 a) or (-3, 1 卤 1/2), which are (-3, 1/2) and (-3, 3/2).
03

Determine the Foci

From our equation, we find that b = 1/4. For a hyperbola, c = \sqrt{a^{2} + b^{2}} and the foci are (h, k 卤 c). Solving for c gives \sqrt{1/4 + 1/16} = \sqrt{5/16} = 5/4, so the foci are (-3, 1 卤 5/4) or (-3, -1/4) and (-3, 9/4).
04

Find the Equations of the Asymptotes

\The equations of the asymptotes are y = k 卤 (a/b)(x - h). Substituting the values we found gives y = 1 卤 (2/4)(x + 3), which simplifies to equations y = 1 + x/2 and y = 1 - x/2 for the upper and lower asymptotes, respectively.
05

Sketch the Hyperbola Using the Asymptotes

The hyperbola should open along the y-axis because its equation has y in the first term. Mark the center at (-3, 1), and plot the vertices (-3, 1/2) and (-3, 3/2) on the y-axis; these represent the 'ends' of the hyperbola. Draw dotted lines for the asymptotes y = 1 + x/2 and y = 1 - x/2, which cross at the center. The hyperbola will approach these lines but never fully reach them, and should fall inside the area defined by them. Use a ruler to sketch the hyperbola as precisely as possible.

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