## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.6 Parametric Equations, Graphs, and Applications - 8.6 Exercises - Page 404: 24

#### Answer

(a) $x = 1+2~sin~t$ $y = 2+3~cos~t$ We can see the graph below. (b) $\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9} = 1$

#### Work Step by Step

(a) $x = 1+2~sin~t$ $y = 2+3~cos~t$ When $t = 0$: $x = 1+2~sin~0 = 1$ $y = 2+3~cos~0 = 5$ When $t = \frac{\pi}{6}$: $x = 1+2~sin~\frac{\pi}{6} = 2$ $y = 2+3~cos~\frac{\pi}{6} = 4.60$ When $t = \frac{\pi}{4}$: $x = 1+2~sin~\frac{\pi}{4} = 2.41$ $y = 2+3~cos~\frac{\pi}{4} = 4.12$ When $t = \frac{\pi}{3}$: $x = 1+2~sin~\frac{\pi}{3} = 2.73$ $y = 2+3~cos~\frac{\pi}{3} = 3.5$ When $t = \frac{\pi}{2}$: $x = 1+2~sin~\frac{\pi}{2} = 3$ $y = 2+3~cos~\frac{\pi}{2} = 2$ When $t = \frac{2\pi}{3}$: $x = 1+2~sin~\frac{2\pi}{3} = 2.73$ $y = 2+3~cos~\frac{2\pi}{3} = 0.5$ When $t = \pi$: $x = 1+2~sin~\pi = 1$ $y = 2+3~cos~\pi = -1$ When $t = \frac{4\pi}{3}$: $x = 1+2~sin~\frac{4\pi}{3} = -0.73$ $y = 2+3~cos~\frac{4\pi}{3} = 0.5$ When $t = \frac{3\pi}{2}$: $x = 1+2~sin~\frac{3\pi}{2} = -1$ $y = 2+3~cos~\frac{3\pi}{2} = 2$ We can see the graph below. (b) $x = 1+2~sin~t$ $\frac{x-1}{2} = sin~t$ $y = 2+3~cos~t$ $\frac{y-2}{3} = cos~t$ $(\frac{x-1}{2})^2+(\frac{y-2}{3})^2 =sin^2~t+cos^2~t$ $\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9} = 1$

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