Question 1125292
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                It is  NOT  AN  ELLIPSE  !      It is a hyperbola ! !


                Fix an error in your condition ! ! ! 



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9x² - 3y² - 36x - 6y + 12 = 0.


Complete the squares for x-terms and y-terms separately, step by step


(9x^2 - 36x) - (3y^2 + 6y) = -12


9*(x^2 - 4x) - 3*(y^2 + 2y) = -12


9*(x^2 -4x + 4) - 3*(y^2 + 2y + 1) = -12 + 9*4 - 3


9*(x-2)^2 - 3*(y+1)^2 = 21.


Divide both sides by 21


{{{(x-2)^2/((21/9))}}} - {{{(y+1)^2/((21/3))}}} = 1,   or equivalently


{{{(x-2)^2/((7/3))}}} - {{{(y+1)^2/7}}} = 1,


{{{(x-2)^2/(sqrt(7/3))^2}}} - {{{(y+1)^2/(sqrt(7))^2}}} = 1.    (Equation of a hyperbola)


Thus   a= {{{sqrt(7/3)}}},  b= {{{sqrt(7)}}},    and  c = {{{sqrt(a^2 + b^2)}}} = {{{sqrt(7/3+7)}}} = {{{sqrt(28/3)}}} = {{{2*sqrt(7/3)}}}.
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