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Chapter 6: Problem 29

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$9 x^{2}-y^{2}-36 x-6 y+18=0$$

Short Answer

Expert verified
The hyperbola has center at (2, -3), vertices at (0, -3) and (4, -3), foci at (2卤sqrt(13), -3), and the equations of the asymptotes are \(y = -3 卤(3/2)(x - 2)\)

Step by step solution

01

Convert to Standard Form

Rearrange and group the x and y terms, and move the constant to the right hand side of the equation:\(9(x^2-4x+4)- (y^2+6y+9) = 18 + 36 + 9. This gives us the equation in standard form as \((x-2)^{2}/4 - (y+3)^{2}/9 = 1\).
02

Find the Center

The center of the hyperbola is given by (h, k), which corresponds to (2, -3) in the standard form.
03

Find the Vertices and Foci

The vertices are given by (h卤a, k), and here a=sqrt(4)=2, so the vertices are at (2卤2, -3), which are the points (0, -3) and (4, -3). The foci are given by (h卤c, k) where \(c=sqrt(a^2 + b^2) = sqrt(4+9) = sqrt(13)\) , so the foci are at (2卤sqrt(13), -3).
04

Find the Equations of the Asymptotes

The equations of the asymptotes are given by \(y = k 卤 (b/a)(x - h)\), substituting the respective values, we get \(y = -3 卤(3/2)(x - 2)\).
05

Sketch the Hyperbola

Using the information from steps 2 to 4, draw the center, vertices, foci, and asymptotes, then sketch the hyperbola based on these points and lines.

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