SOLUTION: what is the standard form of the equation {{{x^2+6y^2-2x+12y-23=0}}}
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Question 586507: what is the standard form of the equation
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You need to complete the square on each of the variables. Group the variables and move the constant term into the RHS:
Divide the coefficient on the 1st degree 
term by 2, square the result, then add that result to both sides:
The coefficient on the 2nd degree 
term is greater than 1, so factor the coefficient from the 1st and second degree terms:
\ =\ 23\ +\ 1)
Divide the coefficient on by 2, square the result, and add that result INSIDE the parentheses. Multiply what you added inside the parentheses by the factor outside the parentheses and add that result to the RHS of the equation.
\ =\ 23\ +\ 1\ +\ 6)
Factor the two perfect square trinomials in the LHS and collect terms in the RHS:
^2\ +\ 6(y\ +\ 1)^2\ =\ 30)
Divide both sides by the constant in the LHS:
^2}{30}\ +\ \frac{(y\ +\ 1)^2}{5}\ =\ 1)
Re-write so that the center coordinates and the measures of the semi-axes can be determined by inspection:
^2}{\left(\sqrt{30}\right)^2}\ +\ \frac{\left(y\ -\ (-1)\right)^2}{\left(\sqrt{5}\right)^2}\ =\ 1)
John
My calculator said it, I believe it, that settles it
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