Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 11 - Section 11.4 - Shifted Conics - 11.4 Exercises - Page 815: 61

Answer

See graph and explanations.

Work Step by Step

Step 1. Rewrite the equation as $9x^2-36x-y^2-6y=-36$ or $9(x^2-4x+4)-(y^2+6y+9)=-36+36-9$ which gives $9(x-2)^2-(y+3)^2=-9$ Step 2. Divide both sides by $-9$, we get $\frac{(y+3)^2}{9}-(x-2)^2=1$ which represents a hyperbola centered at $(2,-3)$ with $a=3, b=1$ Step 3. The above hyperbola can be graphed as shown in the figure.
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