Question 1106320
<br>
(1) complete the squares in both x and y; and
(2) divide by the appropriate constant to get "1" on the right hand side.<br>
{{{-4x^2 + 9y^2 + 32x + 36y - 64 = 0}}}  [original equation]
{{{-4x^2+32x+9y^2+36y = 64}}}  [group the x terms and y terms; move constant to other side]
{{{-4(x^2-8x)+9(y^2+4y) = 64}}}  [factor out the leading coefficients for both x and y]
{{{-4(x^2-8x+16)+9(y^2+4y+4) = 64-4(16)+9(4) = 36}}}  [complete the squares in x and y, adding the same quantities to both sides of the equation]
{{{(-4/36)(x-4)^2+(9/36)(y+2)^2 = 1}}}  [divide by the constant on the right to make the right side "1"]
{{{-(x-4)^2/9+(y+2)^2/4 = 1}}}  [simplify and re-format]
{{{(y+2)^2/4-(x-4)^2/9 = 1}}}  [rearrange terms to get positive term first on left hand side]
{{{(y+2)^2/2^2-(x-4)^2/3^2 = 1}}}  [write in final standard form]<br>
The equation is of a hyperbola with center (4,-2); asymptotes with slopes 2/3 and -2/3; branches opening upward and downward.<br>
A graph: 2 branches of the hyperbola (red, green); asymptotes (blue, purple).  The center is at the intersection of the asymptotes.<br>
{{{graph(400,400,-8,16,-10,10,sqrt(4+(4(x-4)^2/9))-2,-sqrt(4+(4(x-4)^2/9))-2,(2/3)(x-4)-2,(-2/3)(x-4)-2)}}}