Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 10 - Analytic Geometry - 10.4 Summary of the Conic Sections - 10.4 Exercises - Page 997: 43

Answer

$${\text{ellipse}}$$

Work Step by Step

$$\eqalign{ & 4{x^2} - 8x + 9{y^2} - 36y = - 4 \cr & 4\left( {{x^2} - 2x} \right) + 9\left( {{y^2} - 4y} \right) = - 4 \cr & {\text{Complete the square}} \cr & 4\left( {{x^2} - 2x + 1} \right) + 9\left( {{y^2} - 4y + 4} \right) = - 4 + 4\left( 1 \right) + 9\left( 4 \right) \cr & 4{\left( {x - 1} \right)^2} + 9{\left( {y - 2} \right)^2} = 36 \cr & {\text{Divide by 36}} \cr & \frac{{{{\left( {x - 1} \right)}^2}}}{9} + \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 \cr & {\text{The equation is in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr & {\text{Therefore, the equation represents an ellipse}} \cr & {\text{Graph}} \cr} $$
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