College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Conic Sections - Exercise Set 7.1 - Page 671: 55

Answer

$\dfrac{(x+2)^2}{4^2}+\dfrac{(y-3)^2}{8^2}=1$ $(-2,3-4\sqrt 3)$ $(-2,3+4\sqrt 3)$

Work Step by Step

We are given the equation: $4x^2+y^2+16x-6y-39=0$ Form two squares: $(4x^2+16x+16)+(y^2-6y+9)-16-9-39=0$ $4(x^2+4x+4)+(y^2-6y+9)-64=0$ $4(x+2)^2+(y-3)^2=64$ Divide both sides by 64: $\dfrac{4(x+2)^2}{64}+\dfrac{(y-3)^2}{64}=1$ Simplify: $\dfrac{(x+2)^2}{16}+\dfrac{(y-3)^2}{64}=1$ $\dfrac{(x+2)^2}{4^2}+\dfrac{(y-3)^2}{8^2}=1$ Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$: $h=-2$ $k=3$ $a=8$ $b=4$ Determine the foci: $c^2=a^2-b^2$ $c^2=8^2-4^2$ $c^2=48$ $c=\pm\sqrt {48}=\pm 4\sqrt 3$ The foci are: $F_1(h,k-c)=(-2,3-4\sqrt 3)$ $F_2(h,k+c)=(-2,3+4\sqrt 3)$
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