document.write( "Question 580267: What are the foci of the eqation (x+4)^2/9 + (y-2)^2/1 = 1 \n" ); document.write( "
Algebra.Com's Answer #371285 by lwsshak3(11628)\"\" \"About 
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What are the foci of the equation
\n" ); document.write( "(x+4)^2/9 + (y-2)^2/1 = 1
\n" ); document.write( "This is an equation of an ellipse with horizontal axis of the standard form:
\n" ); document.write( "(x-h)^2/a^2+(y-k)^2/b^2=1, a>b, (h,k) being the (x,y) coordinates of the center.
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\n" ); document.write( "For given equation:
\n" ); document.write( "center: (-4,2)
\n" ); document.write( "a^2=9
\n" ); document.write( "b^2=1
\n" ); document.write( "c^2=a^2-b^2=9-1=8
\n" ); document.write( "c=√8
\n" ); document.write( "Foci=(-4±c,2)=(-4±√8,2)=(-4±2.83,2)=(-6.83,2) and (-1.17,2)\r
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