Please observe that the square terms are #5x^2# and #18y^2#; this tells us that we should begin the process of finding the center coordinates #(h,k)# by adding #5h^2+18k^2-27# to both sides of the equation:
#5x^2-30x+5h^2+18y^2+72y+18k^2=5h^2+18k^2-27" [1]"#
Remove a factor of 5 from the first 3 terms on the left and a factor of 18 from the remaining terms on the left:
#5(x^2-6x+h^2)+18(y^2+4y+k^2)=5h^2+18k^2-27" [2]"#
Please observe that the trinomial in the first parenthesis fits the pattern #(x-h)^2= x^2-2hx + h^2#, when:
#-2h = -6#
#h = 3#
Substitute #(x-3)^2# into the trinomial on the left of equation [2] and 3 for h on the right:
#5(x-3)^2+18(y^2+4y+k^2)=5(3)^2+18k^2-27" [3]"#
Please observe that the trinomial in the second parenthesis fits the pattern #(y-k)^2= y^2-2ky + k^2#, when:
#-2k=4#
#k = -2#
Substitute #(y-(-2))^2# into the trinomial on the left of equation [3] and -2 for k on the right:
#5(x-3)^2+18(y-(-2))^2=5(3)^2+18(-2)^2-27#
Simplify the right side:
#5(x-3)^2+18(y-(-2))^2=90#
Divide both sides of the equation by 90:
#(x-3)^2/18+(y-(-2))^2/5=1#
Write the denominator as squares:
#(x-3)^2/(3sqrt2)^2+(y-(-2))^2/(sqrt5)^2=1" [4]"#
Equation [4] fits the standard Cartesian form for the equation of an ellipse with a horizontal major axis:
#(x-h)^2/a^2+(y-k)^2/b^2=1#
where #h=3,k=-2,a=3sqrt2 and b=sqrt5#. From this we know the following:
The center is:
#(h,k) = (3,-2)#
The foci are:
#(h-sqrt(a^2-b^2),k) = (3-sqrt23,-2) and (h+sqrt(a^2-b^2),k) = (3+sqrt23,-2)#
The vertices are:
#(h-a,k) = (3-3sqrt2,-2) and (h+a,k) = (3+3sqrt2,-2)#
The endpoints of the minor axis are:
#(h,k-b) = (3,-2-sqrt5) and (h,k+b) = (3,-2+sqrt5)#
The eccentricity is:
#epsilon = sqrt(1-b^2/a^2) =sqrt(1-5/18) = sqrt(13/18)#
