Question 259707
We have this original equation as
(i) {{{2x^2 + 6y^2 + 32x - 48y + 212 = 0}}}
rearranging, we get
(ii) {{{(2x^2 + 32x) + (6y^2 - 48y) = -212}}}
complete the square to get
(iii) {{{2(x+8)^2 + 6(y-4)^2 = -212 + 128 +96}}}
to get
(iv) {{{2(x+8)^2 + 6(y-4)^2 = 12}}}
dividing by 12, we get
(v) {{{((x+8)^2)/6 + ((y-4)^2)/2 = 1}}}
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center is (-8,4)
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major axis = sqrt(6)
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minor axis = sqrt(2)
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distance to foci
sqrt(sqrt(6) - sqrt(2)) = sqrt(4) = 2