Precalculus (6th Edition)

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

Chapter 10 - Analysis of Variance - Chapter 10 Test Prep - Review Exercises - Page 1000: 19

Answer

$${\text{ellipse}}$$

Work Step by Step

$$\eqalign{ & 4{x^2} - 8x + 9{y^2} + 36y = - 4 \cr & {\text{Factor }} \cr & \left( {4{x^2} - 8x} \right) + \left( {9{y^2} + 36y} \right) = 4 \cr & 4\left( {{x^2} - 2x} \right) + 9\left( {{y^2} + 4y} \right) = 4 \cr & {\text{Complete the square}} \cr & 4\left( {{x^2} - 2x + 1} \right) + 9\left( {{y^2} + 4y + 4} \right) = 4 + 4\left( 1 \right) + 9\left( 4 \right) \cr & 4{\left( {x - 1} \right)^2} + 9{\left( {y + 2} \right)^2} = 44 \cr & {\text{Divide both sides by 44}} \cr & \frac{{{{\left( {x - 1} \right)}^2}}}{{11}} + \frac{{{{\left( {y + 2} \right)}^2}}}{{44/9}} = 1 \cr & {\text{The equation is written in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\,\,\, \cr & {\text{Therefore,}} \cr & {\text{The graph of the equation is an ellipse}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.