document.write( "Question 860695: Find an equation in standard form for the hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
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\n" ); document.write( "The work I have:
\n" ); document.write( "Since c is the distance between the foci and center, c=7-1=6
\n" ); document.write( "c^2=a^2+b^2=36
\n" ); document.write( "Since e=c/a=2, c/a=2/1=6/3
\n" ); document.write( "Since c/a= sqrt.(a^2-b^2)/a, 6/3= sqrt.(9-b^2)/3.
\n" ); document.write( "36=9-b^2
\n" ); document.write( "27=-b^2
\n" ); document.write( "-27=b^2
\n" ); document.write( "So the equation would be [(x-1)^2]/9-[(y+4)^2]/-27.
\n" ); document.write( "but the -27 would change the sign of the y to +
\n" ); document.write( "so it'd be [(x-1)^2]/9+[(y+4)^2]/-27
\n" ); document.write( "But isn't the standard form for hyperbolas [(x-h)^2]/a^2-[(y+k)^2]/b^2?
\n" ); document.write( "How can it be a + and still be a hyperbola?
\n" ); document.write( "Please tell me if I'm doing something wrong. \n" ); document.write( "

Algebra.Com's Answer #518603 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
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document.write( "Hi,
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document.write( "re TY: Foci units units up and down from center
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document.write( "there is NO minus to consider (You may be thinking of the ellipse format)
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document.write( "hyperbola centered at (1, -4), with one focus at (7, -4) and eccentricity e=2.
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document.write( "Center, Vertices and foci along y = -4 
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document.write( "Yes!
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document.write( "2 = c/a and Yes,  c = 6  and a = 3
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document.write( " sqrt(9+ 27) = 6, b^2 = 27   |Foci units units up and down from center
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