Question

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

Answer 1

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]}