Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1022: 99


a) The equation $r=\frac{1}{3-3\cos \theta }$ can be written in the form of $9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$ b) The equation in the rectangular coordinate form is $9{{y}^{2}}=1+6x$ and the curve is a parabola.

Work Step by Step

(a) Let us consider the equation: $r=\frac{1}{3-3\cos \theta }$ We cross multiply the terms: $\begin{align} & r\left( 3-3\cos \theta \right)=1 \\ & 3r-3r\cos \theta =1 \\ & 3r=1+3r\cos \theta \end{align}$ We square both sides of this equation: $\begin{align} & 3r=1+3r\cos \theta \\ & 9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}} \end{align}$ Thus, the equation $r=\frac{1}{3-3\cos \theta }$ can be written in the form of $9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$. (b) Let us consider the equation: $9{{r}^{2}}={{\left( 1+3r\cos \theta \right)}^{2}}$ In polar form, $x=r\cos \theta $ and $y=r\sin \theta $ Therefore, ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ Then, put these values in the equation $\begin{align} & 9\left( {{x}^{2}}+{{y}^{2}} \right)={{\left( 1+3x \right)}^{2}} \\ & 9{{x}^{2}}+9{{y}^{2}}=1+9{{x}^{2}}+6x \\ & 9{{y}^{2}}=1+6x \end{align}$ Thus, the provided equation is a parabola opening towards the right.
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