Question

How do you find the standard form of `9x^2-16y^2-90x+32y+65=0`?

Answer 1

`(x-5)^2/4^2-(y-1)^2/3^2=1`

Explanation:

`9x^2-90x=9(x-5)^2-225`.
`-19y^2+32y=-16(y-1)^2+16`.

So, the given equation becomes
`9(x-5)^2-16(y-1)^2=144`
Divide by 144.
`(x-5)^2/4^2-(y-1)^2/3^2=1`.

This equation represents a hyper bola with semi-major axis a = 4 and semi-transverse axis b = 3.

The eccentricity of the hyperbola is 1.25, using the relation `b^2=a^2(e^2-1)`

The asymptotes are `y=+-(b/a)x=+-3/4x`.