Find the equation, center, foci, and the lengths of the major & minor axes from #x^2 +2y^2 -12y +4x+2=0#?

1 Answer
Jun 26, 2017

#(x - (-2))^2/(2sqrt5)^2 +(y-3)^2/(sqrt(10))^2 =1#

Center: #(-2,3)#

Foci: #(-2-sqrt(10),3)# and #(-2+sqrt(10),3)#

Major axis: #4sqrt(5)#

Minor axis: #2sqrt10#

Explanation:

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#