Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Chapter Review - Page 672: 31


See below

Work Step by Step

Given: $4x^2+9y^2+40x+72y+208=0$ Because $A=4,B=0$, and $C=9$, the discriminant is $B^2-4AC=-144\lt0$, so the conic is an ellipse. Complete the square to write the equation in standard form. $4x^2+9y^2+40x+72y+208=0\\4(x^2+10x+25)-100+9(y^2+8y+16)-144+208=0\\4(x+5)^2+9(y+4)^2=36\\\frac{(x+5)^2}{9}+\frac{(y+4)^2}{4}=1$ From the equation, we can see that $(-5,-4)$ and $a=3,b=2$. The graph is shown below.
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