Question 736460
write this hyperbola in standard form
{{{-16x^2+4y^2-64x+24y+36=0}}}
rearrange terms
{{{-16x^2-64x+4y^2+24y+36=0}}}
complete the square
{{{-16(x^2+4x+4)+4(y^2+6y+9)=-36-64+36}}}
{{{-16(x+2)^2+4(y+3)^2=-64}}}
divide by -64
{{{(x+2)^2/4-(y+3)^2/16=1}}}
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form:{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center