To demonstrate, let's try transforming the equation below into standard form.
4x^2 + 9y^2 + 72y - 24x + 144 = 04x2+9y2+72y−24x+144=0
The first thing to do is group xxs and yys together
(4x^2 - 24x) + (9y^2 + 72y) + 144(4x2−24x)+(9y2+72y)+144
Factor out x^2x2's and y^2y2's coefficient
4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 04(x2−6x)+9(y2+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(ax+b)2=a2x2+2abx+b2
For our problem, we want x^2 - 6xx2−6x and y^2 + 8yy2+8y to be perfect squares
For xx, we know that
a^2 = 1a2=1
=> a = 1, a = -1⇒a=1,a=−1
2ab = -62ab=−6
2(1)b = -62(1)b=−6
2b = -62b=−6
b = -3b=−3
=> b^2 = 9⇒b2=9
Note that substituting a = -1a=−1 will result to the same b^2b2
Hence, to complete the square, we need to add 99
Meanwhile, for yy
a^2 = 1a2=1
=> a = 1, a = -1⇒a=1,a=−1
2ab = 82ab=8
2(1)b = 82(1)b=8
2b = 82b=8
b = 4b=4
=> b = 16⇒b=16
Now, let's add our b^2b2s into the equation.
4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 04(x2−6x)+9(y2+8y)+144=0
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 04(x2−6x+9)+9(y2+8y+16)+144=0
We added something into the left-hand side, to retain the "equality", we need to add the same value to the right-hand side (or substract the value again from the left-hand side)
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0 + 4(9) + 9(16)4(x2−6x+9)+9(y2+8y+16)+144=0+4(9)+9(16)
4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 1804(x2−6x+9)+9(y2+8y+16)+144=180
=> 4(x - 3)^2 + 9(y + 4)^2 + 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=180−144
=> 4(x - 3)^2 + 9(y + 4)^2 = 36⇒4(x−3)2+9(y+4)2=36
=> (4(x - 3)^2 + 9(y + 4)^2 = 36)/36⇒4(x−3)2+9(y+4)2=3636
=> (x - 3)^2/9 + (y + 4)^2/4 = 1⇒(x−3)29+(y+4)24=1
