Question

How do I graph `-9x^2+25y^2+36x+150y-36=0` algebraically?

Answer 1

`-9x^2 + 36x + 25y^2 + 150y = 36`
`-9(x^2 - 4x) + 25(y^2 + 6y) = 36`
`-9(x^2 - 4x color(red) ( + 4)) + 25(y^2 + 6y color(red)(+9)) = 36 color(red)(-36+225)`
`-9(x-2)^2 +25(y+3)^2 = 225`
`((-9(x-2)^2)/225) + ((25(y+3)^2)/225) = 1`

`-(x-2)^2/25 + (y+3)^2/9 = 1`

`(y+3)^2/9 - (x-2)^2/25 = 1`

You now have an equation of a hyperbola with vertex at `(2, -3)`, extending vertically with asymptotes at a slope of `+-3/5.`

Asymptotes: `y = +-3/5(x-2)-3`
The graph of the function:
graph{(y+3)^2/9 -(x-2)^2/25 = 1 [-10, 10, -5, 5]}