Question

an someone explain me how you write the equation of a hyperbola in standard (completed square) form:
(x)^2+4(y)^2+6x-8y+9=0

Answer 1

Thanks i got the answer already :DDD

Answer 2

You have got to separate the x terms and y terms. Then make a whole square of each term. Just see:
\[x^{2}+6x+4y ^{2}-8y+9=0\]
\[x^{2}+2*3x+4y ^{2}-8y+9=0\]
\[x^{2}+2*3x+9+4y ^{2}-8y=0\]
\[(x+3)^{2}+4(y ^{2}-2y)=0\]
\[(x+3)^{2}+4(y ^{2}-2y+1)=4\]
\[(x+3)^{2}+4(y-1)^{2}=4\]
Are u sure this is a hyperbola's equation? It turns out to be an ellipse's.

Answer 3

Oh sorry i didnt see that u already got it.