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Question 30414: how would you solve for the center, major axis, and minor axis of the following problem: x^2+4y^2+24y=-32
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! x^2+4y^2+24y=-32----(1)
x^2+4(y^2+6y) = -32
x^2+4[(Y+3)^2-9]=-32
x^2+4(y+3)^2-36 = -32
x^2+4(y+3)^2= -32+36
x^2+4(y+3)^2 = 4
Dividing by 4
x^2/4 + (y+3)^2/1 = 1
That is [(x-0)^2]/(2^2) +[y-(-3)]^2/(1^2) = 1 ----(1)
(since 2^2 > 1^2 and 2^2 as dr under [(x-0)^2],
this means the major axis is horizontal
(1) is of the general form (x-h)^2/a^2 +(y-k)^2/b^2 = 1 ----(I)
Comparing our (1) with the general ellipse (I)
we have centre, C = (h,k) = (0,-3)
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Draw the horizontal and vertical through C(0,-3)
The horizontal through C(0,-3) is the major axis.
and the equation to the major axis is given by y = -3
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The vertical through C(0,-3), that is the y-axis is the minor axis.
and the equation to the minor axis is given by x = 0
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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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