Given:
#x^2 + 8y^2 − 8x − 16y − 40 = 0#
Add 40 to both sides:
#x^2 + 8y^2 − 8x − 16y = 40#
Group the x terms and the y terms together:
#(x^2 − 8x) + (8y^2 − 16y) = 40#
We cannot complete the square unless the leading coefficient is 1, therefore, we remover a factor of 8 from the y terms:
#(x^2 − 8x) + 8(y^2 − 2y) = 40#
Because #(x-h)^2= x^2-2hx+h^2# we want to insert an #h^2# into the x term's group but we must, also add #h^2# to the right side so that equality is maintained:
#(x^2 − 8x+h^2) + 8(y^2 − 2y) = 40+h^2#
Matching the x terms with the general pattern, #(x-h)^2= x^2-2hx+h^2#, we observe that the equation
#-2hx = -8x#
will allow us to solve for the value of h:
#h = 4#
This means that #h^2# on the right side becomes 16 and the group of x terms become #(x-4)^2#
#(x − 4)^2 + 8(y^2 − 2y) = 40+16#
Combine like terms:
#(x − 4)^2 + 8(y^2 − 2y) = 56#
We want insert #k^2# into the y terms but to maintain equality we must add #8k^2# to the right side:
#(x − 4)^2 + 8(y^2 − 2y+ k^2) = 56+8k^2#
Matching the y terms with the general pattern, #(y-k)^2= y^2-2ky+k^2#, we observe that the equation
#-2ky = -2y#
will allow us to solve for the value of k:
#k = 1#
This means that #8k^2# on the right side becomes 8 and the group of y terms become #(y-1)^2#
#(x − 4)^2 + 8(y − 1)^2 = 56+8#
Combine like terms:
#(x − 4)^2 + 8(y − 1)^2 = 64#
Divide both sides by 64:
#(x − 4)^2/64 + (y − 1)^2/8 = 1#
Write the denominators as squares:
#(x − 4)^2/8^2 + (y − 1)^2/(2sqrt2)^2 = 1#
This is the same form as equation [1],
#(x-h)^2/a^2+(y-k)^2/b^2=1" [1]"#
The center, #(h,k) = (4,1)#
The vertices are, #(h-a,k) = (-4,1)# and #(h+a,k)= (12,1)#
The foci are, #(h-sqrt(a^2-b^2),k) = (4-sqrt56,1)# and #(h+sqrt(a^2-b^2),k) = (4+sqrt56,1)#
The eccentricity is #sqrt(a^2-b^2)/a = sqrt56/8#
