Question 30414
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)
------------------------------------
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
-----------------------------------------------------------
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
-----------------------------------------------------------

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,