Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - 10.2 The Parabola - 10.2 Assess Your Understanding - Page 647: 50


Vertex: $(-3,-2)$ Focus: $\left(-3,-1\right)$ Directrix: $y=-3$ See graph

Work Step by Step

We are given the parabola: $x^2+6x-4y+1=0$ Put the equation in standard form: $(x^2+6x+9)-9-4y+1=0$ $(x+3)^2=4y+8$ $(x+3)^2=4(y+2)$ The standard equation is: $(x-h)^2=4p(y-k)$ Determine $h,k,p$: $h=-3$ $k=-2$ $4p=4\Rightarrow p=1$ Determine the vertex: $(h,k)=(-3,-2)$ Determine the focus: $(h,k+p)=\left(-3,-2+1\right)=\left(-3,-1\right)$ Determine the directrix: $y=k-p$ $y=-2-1$ $y=-3$ Determine the two points defining the latus rectum: $y=-1$ $(x+3)^2=4(-1+2)$ $(x+3)^2=4$ $x+3=\pm 2$ $x+3=-2\Rightarrow x_1=-5$ $x+3=2\Rightarrow x_2=-1$ $\Rightarrow (-5,-1),(-1,-1)$ Plot the points, draw the directrix, and graph the parabola:
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