Question

How do you write the standard form of the hyperbola `-16x^2+9y^2+32x+144y-16=0`?

Answer 1

We can rewrite this as follows

`16x^2-9y^2-32x-144y+16=0` (Multiply with `-1` both sides)

`16*(x^2-2x)-9(y^2+16y)+16=0`

`16*(x^2-2x+1)-9(y^2+16+64)+(16-16+576)=0`

`16*(x-1)^2-9*(y+8)^2=-576`

`(y+8)^2/64-(x-1)^2/36=1`

`((y+8)/8)^2-((x-1)/6)^2=1`

Finally the standard form for the hyperbola is

`(y-(-8))^2/8^2-(x-1)^2/6^2=1`

If we plot it on the cartesian plane it is