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#
