document.write( "Question 188598: Give the vertex, focus, directrix, and the distance across parabola at focus, and then graph the parabola.
\n" ); document.write( "y² - 8x - 4y + 12 = 0 \n" ); document.write( "
Algebra.Com's Answer #141494 by nerdybill(7384)\"\" \"About 
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y^2-8x-4y+12=0
\n" ); document.write( "Isolate the 'x' to one side of the equation:
\n" ); document.write( "y^2-4y+12 = 8x
\n" ); document.write( "(1/8)y^2-(1/2)y+(3/2) = x
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\n" ); document.write( "Because 'y' is squared it will be a HORIZONTAL parabola.
\n" ); document.write( "Because the coefficient associated with the y^2 term is positive: open right
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\n" ); document.write( "Completing the square:
\n" ); document.write( "(1/8)(y^2-4y)+(3/2) = x
\n" ); document.write( "(1/8)(y^2-4y+4)+(3/2 - 1/2) = x
\n" ); document.write( "(1/8)(y^2-4y+4) + 1 = x
\n" ); document.write( "(1/8)(y-2)^2 + 1 = x
\n" ); document.write( ".
\n" ); document.write( "This now is in the form of:
\n" ); document.write( "x = (1/(4c))(y-k)^2 + h
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\n" ); document.write( "From the above, we see that
\n" ); document.write( "vertex = (1,2)
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\n" ); document.write( "c = distance between vertex and the focus/directrix
\n" ); document.write( "1/(4c) = 1/8
\n" ); document.write( "cross-multiplying:
\n" ); document.write( "4c = 8
\n" ); document.write( "c = 2
\n" ); document.write( ".
\n" ); document.write( "focus is at (h+c,k)
\n" ); document.write( "focus = (1+2, 2) = (3, 2)
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\n" ); document.write( "directrix is at:
\n" ); document.write( "x = h-c
\n" ); document.write( "x = 1-2
\n" ); document.write( "x = -1
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