Question

Identify the standard form of the equation by completing the square

Answer 1

$4 x^{2} -9 y^{2} -8x-36y-68=0$

$4 x^{2} -8x-(9 y^{2} +36y)-68=0$

$(2x)^{2} -2*(2x)*2-((3y)^{2} +2*(3y)*6)-68=0$

add and subtract 2^ to the x-expression and 6^ to the second expression to complete the square.

$[(2x)^{2} -2*(2x)*2+4]-4-((3y)^{2} +2*(3y)*6+36-36)-68=0$

$(2x-2)^{2} -4-((3y+6)^{2} -36)-68=0$

$(2x-2)^{2} -(3y+6)^{2} -4+36-68=0$

$(2x-2)^{2} -(3y+6)^{2} -36=0$

$(2x-2)^{2} -(3y+6)^{2} = 6^{2}$

divide by 6 squared 

$( \frac{2x-2 }{6} )^{2} -( \frac{3y+6}{6} )^{2} =1$

$( \frac{x-1 }{3} )^{2} -( \frac{y+2}{2} )^{2} =1$

$\frac{ (x-1)^{2} }{9} - \frac{(y+2)^{2}}{4} =1$

Answer: A