# Chapter 10 - Analytic Geometry - 10.6 Polar Equations of Conics - 10.6 Assess Your Understanding - Page 685: 32

$12x^2-4y^2+64-64x=0$

#### Work Step by Step

Cross-multiply to obtain: $r(2+4\cos{\theta})=8\\2r+4r\cos{\theta}=8\\2r=8-4r\cos{\theta}$ Square both sides: $(2r)^2=\left(8-4r\cos{\theta}\right)^2\\ \\4r^2=(8-4r\cos{\theta})^2$. We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$. Hence $4r^2=(8-4r\cos{\theta})^2$ becomes: \begin{align*} 4(x^2+y^2)&=(8-4x)^2\\ 4x^2+4y^2&=64-64x+16x^2\\ 0&=64-64x+16x^2-4x^2-4y^2\\ 0&=12x^2-4y^2-64x+64 \end{align*}

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