Answer
$\frac{(y-4)^2}{9}-\frac{(x-2)^2}{9}=1$
Work Step by Step
We know that $\tan^2 A+1=\sec^2 A$. $\tan t=\frac{x-2}{3},\sec t=\frac{y-4}{3}$
Hence here $(\frac{x-2}{3})^2+1=(\frac{y-4}{3})^2$, thus $\frac{(y-4)^2}{9}-\frac{(x-2)^2}{9}=1$
Trigonometry 7th Edition
by McKeague, Charles P.; Turner, Mark D.
Published by
Cengage Learning
ISBN 10:
1111826854
ISBN 13:
978-1-11182-685-7
Chapter 6 - Section 6.4 - Parametric Equations and Further Graphing - 6.4 Problem Set - Page 352: 33
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
