document.write( "Question 30414: how would you solve for the center, major axis, and minor axis of the following problem: x^2+4y^2+24y=-32 \n" ); document.write( "
Algebra.Com's Answer #17091 by sdmmadam@yahoo.com(530)\"\" \"About 
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x^2+4y^2+24y=-32----(1)
\n" ); document.write( "x^2+4(y^2+6y) = -32
\n" ); document.write( "x^2+4[(Y+3)^2-9]=-32
\n" ); document.write( "x^2+4(y+3)^2-36 = -32
\n" ); document.write( "x^2+4(y+3)^2= -32+36
\n" ); document.write( "x^2+4(y+3)^2 = 4
\n" ); document.write( "Dividing by 4
\n" ); document.write( "x^2/4 + (y+3)^2/1 = 1
\n" ); document.write( "That is [(x-0)^2]/(2^2) +[y-(-3)]^2/(1^2) = 1 ----(1)
\n" ); document.write( "(since 2^2 > 1^2 and 2^2 as dr under [(x-0)^2],
\n" ); document.write( "this means the major axis is horizontal
\n" ); document.write( "(1) is of the general form (x-h)^2/a^2 +(y-k)^2/b^2 = 1 ----(I)
\n" ); document.write( "Comparing our (1) with the general ellipse (I)
\n" ); document.write( "we have centre, C = (h,k) = (0,-3)
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\n" ); document.write( "Draw the horizontal and vertical through C(0,-3)
\n" ); document.write( "The horizontal through C(0,-3) is the major axis.
\n" ); document.write( " and the equation to the major axis is given by y = -3
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\n" ); document.write( "The vertical through C(0,-3), that is the y-axis is the minor axis.
\n" ); document.write( " and the equation to the minor axis is given by x = 0
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\n" ); document.write( "\n" ); document.write( "Caution: When you perfect the square in y, the perfecting is done within the brackets and hence the addition and subtraction of 9 should be within the brackets,
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