Calculus (3rd Edition)

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

Chapter 13 - Vector Geometry - 13.6 A Survey of Quadratic Surfaces - Exercises - Page 693: 44


$$ x^2+y^2=\left(\frac{ka}{c}\right)^{2} +a^2,$$ $$ x^2+y^2=\left(\frac{ka}{c}\right)^{2} -a^2.$$

Work Step by Step

The equation of a hyperboloid of one sheet is $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1+\left(\frac{z}{c}\right)^{2} .$$ To find the horizontal traces, we put $ z=k $; hence the equation becomes $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1+\left(\frac{k}{c}\right)^{2} .$$ For this equation to be a circle, we must have $ a=b $ and hence $$ x^2+y^2=a^2+\left(\frac{ka}{c}\right)^{2} .$$ Similarly, for the hyperboloid of two sheets, we get $$ x^2+y^2=\left(\frac{ka}{c}\right)^{2} -a^2.$$
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.