We are given the General Cartesian Form of a conic section:
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
Where A = 1, B = 0, C = 2, D = 4, E = -12 and F= 2
x^2 +2y^2 -12y +4x+2=0" [1]"
The reference tells us that we can determine that the the equation is an ellipse by observing:
B^2-4AC = 0-4(1)(2) = -8 < 0
Therefore, we shall strive to convert equation [1] into the one of two standard Cartesian forms for an ellipse:
(x-h)^2/a^2+(y-k)^2/b^2=1; a>b" [2]"
OR
(y-k)^2/a^2+(x-h)^2/b^2+=1; a>b" [3]"
In equation [1], move the constant term to the right and group the x terms and y terms together:
x^2+4x +2y^2 -12y =-2" [1.1]"
Add h^2+2k^2 to both sides:
x^2+4x+ h^2 +2y^2 -12y+ 2k^2 =-2+ h^2+2k^2" [1.2]"
Remove a common factor of 2 from the y terms:
x^2+4x+ h^2 +2(y^2 -6y+ k^2) =-2+ h^2+2k^2" [1.3]"
From the pattern (x-h)^2 = x^2-2hx+h^2, please observe that we can find the value of h by setting -2hx equal to 4x:
-2hx = 4x
h = -2
Into equation [1.3], we can substitute (x - (-2))^2 for x^2+4x+ h^2 and 4 for h^2:
(x - (-2))^2 +2(y^2 -6y+ k^2) =-2+ 4+2k^2" [1.4]"
From the pattern (y-k)^2 = y^2-2ky+k^2, please observe that we can find the value of h by setting -2ky equal to -6y:
-2ky = -6y
k = 3
We can substitute (y-3)^2 for y^2 -6y+ k^2 and 18 for 2k^2, into equation [1.4]
(x - (-2))^2 +2(y-3)^2 =-2+ 4+18" [1.5]"
Combine the constant terms on the right:
(x - (-2))^2 +2(y-3)^2 =20" [1.6]"
Divide both sides of the equation by 20:
(x - (-2))^2/20 +(y-3)^2/10 =1" [1.7]"
Convert the denominators to squares:
(x - (-2))^2/(2sqrt5)^2 +(y-3)^2/(sqrt(10))^2 =1" [1.8]"
From equation [1.8], we can read:
center (h,k) = (-2,3)
foci: (h-sqrt(a^2-b^2),k) and (h-sqrt(a^2-b^2),k) =
(-2-sqrt(20-10),3) and (-2+sqrt(20-10),3)=
(-2-sqrt(10),3) and (-2+sqrt(10),3)
Major axis: 2a = 4sqrt(5)
Minor axis: 2b = 2sqrt10
