Question 639080
Convert this general equation of a hyperbola to standard form.
-4x^2+y^2-8x-12y+36=0
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-4x^2+y^2-8x-12y+36=0
-4x^2-8x+y^2-12y+36=0
complete the square
-4(x^2+2x+1)+(y^2-12y+36)=-36-4+36
-4(x+1)^2+(y-6)^2=-4
divide by -4
standard form of equation:{{{(x+1)^2-(y-6)^2/4=1}}}
This is an equation of a hyperbola with horizontal transverse axis (x-term leading)
Its standard generic form: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
center: (-1,6)
a^2=1
b^2=4