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: 61

Answer

See graph $(0,-2),(1,0)$

Work Step by Step

We are given the equations: $\begin{cases} 4x^2+y^2=4\\ 2x-y=2 \end{cases}$ $\begin{cases} \dfrac{4x^2}{4}+\dfrac{y^2}{4}=1\\ y=2x-2 \end{cases}$ $\begin{cases} \dfrac{x^2}{1}+\dfrac{y^2}{4}=1\\ y=2x-2 \end{cases}$ The first equation represents an ellipse. Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$: $h=0$ $k=0$ $a^2=4\Rightarrow a=2$ $b^2=1\Rightarrow b=1$ Determine the foci: $c^2=a^2-b^2$ $c^2=2^2-1^2$ $c^2=3$ $c=\pm\sqrt 3$ The foci are: $F_1(h,k-c)=(0,-\sqrt 3)$ $F_2(h,k+c)=(0,\sqrt 3)$ The second equation represents a line with slope 2 and $y$-intersection $2$. Draw the two equations. There are 2 intersections: $(0,-2),(1,0)$
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