Standard Form of the Equation
Key Questions

Locate the larger of the 2 denominators. Then look at that numerator.
If it has an
#x# then that is the major axis. Horizontal
If it has an#y# then that is the major axis. Vertical#x^2/4+y^2/16=1# The major axis is
#y# . Vertical#x^2/25+y^2/16=1# The major axis is
#x# . Horizontal 
The standard form of the ellipse, centered in the point
#C(x_C,y_C)# and with the semiaxes#a# , horizontal and#b# , vertical is:#(xx_C)^2/a^2+(yy_C)^2/b^2=1# . 
For ellipses,
#a >= b# (when#a = b# , we have a circle)#a# represents half the length of the major axis while#b# represents half the length of the minor axis.This means that the endpoints of the ellipse's major axis are
#a# units (horizontally or vertically) from the center#(h, k)# while the endpoints of the ellipse's minor axis are#b# units (vertically or horizontally)) from the center.The ellipse's foci can also be obtained from
#a# and#b# .
An ellipse's foci are#f# units (along the major axis) from the ellipse's centerwhere
#f^2 = a^2  b^2#
Example 1:
#x^2/9 + y^2/25 = 1# #a = 5#
#b = 3# #(h, k) = (0, 0)# Since
#a# is under#y# , the major axis is vertical.So the endpoints of the major axis are
#(0, 5)# and#(0, 5)# while the endpoints of the minor axis are
#(3, 0)# and#(3, 0)# the distance of the ellipse's foci from the center is
#f^2 = a^2  b^2# #=> f^2 = 25  9#
#=> f^2 = 16#
#=> f = 4# Therefore, the ellipse's foci are at
#(0, 4)# and#(0, 4)#
Example 2:
#x^2/289 + y^2/225 = 1# #x^2/17^2 + y^2/15^2 = 1# #=> a = 17, b = 15# The center
#(h, k)# is still at (0, 0).
Since#a# is under#x# this time, the major axis is horizontal.The endpoints of the ellipse's major axis are at
#(17, 0)# and#(17, 0)# .The endpoints of the ellipse's minor axis are at
#(0, 15)# and#(0, 15)# The distance of any focus from the center is
#f^2 = a^2  b^2#
#=> f^2 = 289  225#
#=> f^2 = 64#
#=> f = 8# Hence, the ellipse's foci are at
#(8, 0)# and#(8, 0)# 
To demonstrate, let's try transforming the equation below into standard form.
#4x^2 + 9y^2 + 72y  24x + 144 = 0# The first thing to do is group
#x# s and#y# s together#(4x^2  24x) + (9y^2 + 72y) + 144# Factor out
#x^2# 's and#y^2# 's coefficient#4(x^2  6x) + 9(y^2 + 8y) + 144 = 0# Before we continue, let's recall what happens when a binomial is squared
#(ax + b)^2 = a^2x^2 + 2abx + b^2#
For our problem, we want
#x^2  6x# and#y^2 + 8y# to be perfect squaresFor
#x# , we know that#a^2 = 1#
#=> a = 1, a = 1# #2ab = 6#
#2(1)b = 6#
#2b = 6#
#b = 3#
#=> b^2 = 9# Note that substituting
#a = 1# will result to the same#b^2# Hence, to complete the square, we need to add
#9#
Meanwhile, for
#y# #a^2 = 1#
#=> a = 1, a = 1# #2ab = 8#
#2(1)b = 8#
#2b = 8#
#b = 4#
#=> b = 16#
Now, let's add our
#b^2# s into the equation.#4(x^2  6x) + 9(y^2 + 8y) + 144 = 0# #4(x^2  6x + 9) + 9(y^2 + 8y + 16) + 144 = 0# We added something into the lefthand side, to retain the "equality", we need to add the same value to the righthand side (or substract the value again from the lefthand side)
#4(x^2  6x + 9) + 9(y^2 + 8y + 16) + 144 = 0 + 4(9) + 9(16)#
#4(x^2  6x + 9) + 9(y^2 + 8y + 16) + 144 = 180# #=> 4(x  3)^2 + 9(y + 4)^2 + 144 = 180#
#=> 4(x  3)^2 + 9(y + 4)^2 = 180  144#
#=> 4(x  3)^2 + 9(y + 4)^2 = 36# #=> (4(x  3)^2 + 9(y + 4)^2 = 36)/36# #=> (x  3)^2/9 + (y + 4)^2/4 = 1#