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 - 9.7 Solve Quadratic Systems - 9.7 Exercises - Quiz for Lessons 9.6-9.7 - Page 664: 7

Answer

See below

Work Step by Step

Given $12x^2+45y^2+120x+90y-150=0$ We can see that $a=9\\b=0\\c=-4$ We will find the discriminant of the given equation $=b^2-4ac\\=0^2-4(9)(-4)\\=-2160$ Since the discriminant $-2160\lt0$, the conic is an ellipse. To graph the ellipse, first complete the square in x. $12x^2+45y^2+120x+90y-150=0\\12(x^2+10x+25)-300+45(y^2+2y+1)-45+7=150\\12(x+5)^2+45(y+1)^2=495\\\frac{(x+5)^2}{41.25}+\frac{(y+1)^2}{11}=1$ From the equation, you can see that the center is at $(-5,-1)$
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.