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Chapter 6: Problem 26

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{(x+3)^{2}}{144}-\frac{(y-2)^{2}}{25}=1$$

Short Answer

Expert verified
The center of the hyperbola is at (-3, 2), its vertices are at (-15, 2) and (9, 2) and its co-vertices are at (-3, -3) and (-3, 7). The foci are at (-16, 2) and (10, 2). The equations of the asymptotes are \(y = (5/12)x + 3/4\) and \(y = -(5/12)x + 5/4\).

Step by step solution

01

Identify the Center, Vertices, and Co-vertices

The equation is in the form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\). Comparing this with \((x+3)^2/144 - (y-2)^2/25 = 1\), obtains h = -3 and k = 2, and thus the center of the hyperbola is (-3, 2). Identifying a^2 = 144 and b^2 = 25, so a = 12 and b = 5. Then calculate the vertices: these are a units left and right of the center, therefore at positions (-3 卤 12, 2), or (-15, 2) and (9, 2). The co-vertices are b units above and below the center, at positions (-3, 2 卤 5), which means (-3, -3) and (-3, 7).
02

Determine the Foci

The foci are found at points a distance of c units away from the center, where c = \(\sqrt{a^2 + b^2}\) = \(\sqrt{144 + 25}\) = \(\sqrt{169}\) = 13. Therefore, the foci are located at (-3 卤 13, 2), or (-16, 2) and (10, 2).
03

Find the Equations of the Asymptotes

The asymptotes follow the equation \(y = k 卤 (b/a)(x - h)\). Substituting the known values gives \(y = 2 卤 (5/12)(x + 3)\), simplifying to \(y = 2 卤 (5/12)x -5/4\), or in complete form as \(y = (5/12)x + 3/4\) and \(y = -(5/12)x + 5/4\).
04

Draw the Hyperbola and Asymptotes

Draw the vertical and horizontal asymptotes as lines crossing at the center of the hyperbola. The hyperbola approaches these asymptotes but never crosses them. Mark the vertices and foci on the x-axis and make the curves of the hyperbola smooth and symmetrical about both the vertical and horizontal centerlines.

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