Precalculus (10th Edition)

Published by Pearson

Chapter 10 - Analytic Geometry - 10.6 Polar Equations of Conics - 10.6 Assess Your Understanding - Page 684: 5

True

Work Step by Step

We are given the equation in polar coordinates: $r=\dfrac{2}{2+3\sin\theta}$ Rewrite the equation and substitute $r$ and $\sin\theta$ using the formulas: $r=\sqrt{x^2+y^2}$ $y=r\sin\theta$ $r(2+3\sin\theta)=2$ $2r+3r\sin\theta=2$ $2r=2-3r\sin\theta$ $(2r)^2=(2-3r\sin\theta)^2$ $4r^2=(2-3r\sin\theta)^2$ $4(x^2+y^2)=(2-3y)^2$ $4x^2+4y^2=4-12y+9y^2$ $4x^2-5y^2+12y-4=0$ Determine the coefficients $A,B,C$ from the general equation: $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ $A=4$ $B=0$ $C=-5$ Compute $B^2-4AC$: $B^2-4AC=0^2-4(4)(-5)=80>0$ Because $B^2-4AC>0$, the equation represents a hyperbola. Therefore the statement is TRUE.

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