How do you identify the conic section represented by the equation #9y^2-16x^2-72y-64x=64#?

1 Answer
Dec 3, 2016

Answer:

This is the equation of a hyperbola #(y-4)^2/16-(x+2)^2/9=1#

Explanation:

Let rearrange the equation and complete the squares

#9y^2-72y-16x^2-64x=64#

#9(y^2-8y)-16(x^2+4x)=64#

#9(y^2-8y+16)-16(x^2+4x+4)=64+144-64#

#9(y-4)^2-16(x+2)^2=144#

#9/144(y-4)^2-16/144(x+2)^2=1#

#(y-4)^2/16-(x+2)^2/9=1#

This is a hyperbola,

#(y-h)^2/a^2-(y-k)^2/b^2=1#

The center is #=(-2,4)#

The equations of the asymptotes are

#y-4=4/3(x+2)#

and #y-4=-4/3(x+2)#

graph{(9y^2-16x^2-72y-64x-64)(y-4-4/3(x+2))(y-4+4/3(x+2))=0 [-83.36, 83.36, -41.6, 41.7]}