Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 43

Answer

The equation has no real solutions. Hence, it has no points.

Work Step by Step

Write ${x^2} + 3{y^2} - 6x + 12y + 23 = 0$ ${x^2} - 6x + 3{y^2} + 12y + 23 = 0$ ${\left( {x - 3} \right)^2} - 9 + 3{\left( {y + 2} \right)^2} - 12 + 23 = 0$ ${\left( {x - 3} \right)^2} + 3{\left( {y + 2} \right)^2} + 2 = 0$ ${\left( {x - 3} \right)^2} + 3{\left( {y + 2} \right)^2} = - 2$ The left-hand side is nonnegative because it is the sum of two squares. But the right-hand side is negative, so there is no real solutions. Hence, it has no points.
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.