SOLUTION: how do i put this hyperbola in to standard form of ((x-h)^2/a^2)-((y-k)^2/b^2)=1 by completing the square 36x^2-y^2-24x+6y-91=0 i know i have to group the x's and y's but when i

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: how do i put this hyperbola in to standard form of ((x-h)^2/a^2)-((y-k)^2/b^2)=1 by completing the square 36x^2-y^2-24x+6y-91=0 i know i have to group the x's and y's but when i       Log On


   

Question 760858: how do i put this hyperbola in to standard form of ((x-h)^2/a^2)-((y-k)^2/b^2)=1 by completing the square
36x^2-y^2-24x+6y-91=0
i know i have to group the x's and y's but when i group the x's (36x^2-24x) how do i make 36x^2 just x^2 because i factored out 12 to get 12(3x^2-2x) but the 3 in from of the x^2 is really confusing me please help :( asap Thank youuu

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
36x%5E2-y%5E2-24x%2B6y-91=0

36x%5E2-24x-y%5E2%2B6y=91
%2836x%5E2-24x%29-%28y%5E2-6y%29=91
Now DIVIDE the x expression by 36 and indicate a factor of 36 applied to the x expression.
36%28x%5E2-%2824%2F36%29x%29-%28y%5E2-6y%29=91
36%28x%5E2-%282%2F3%29x%29-%28y%5E2-6y%29=91

The missing square term for the x part is %282%2F%282%2A3%29%29%5E2=1%2F9
and
The missing square term for the y part is %286%2F2%29%5E3=9

Watch the signs and factors very carefully when you ADD the square terms to both sides of the equation, because this can be the source of mistakes.