Question

(3x^2)-(2y^2)-9x+4y-8=0 Graph and find all applicable points (center, vertex, focus, asymptote)?

I got the equation to be `(x-3/2)^2/(12/67)-(y+1)^2/(8/67)`=1

I got crazy fractions for vertex and a focus that isn't even inside the curves, so I'm pretty sure my answer is incorrect.

Answer 1

Nothing crazy about fractions you got. But see the difference with what I got.

Explanation:

graph{(3x^2)-(2y^2)-9x+4y-8=0 [-10, 10, -4, 6]}
`(3x^2)-(2y^2)-9x+4y-8=0`
Paring `x and y` terms
`(3x^2-9x)-(2y^2-4y)-8=0`
Rearranging
`3(x^2-3x)-2(y^2-2y)-8=0`
`=>3(x^2-2. 3/2x)-2(y^2-2y)-8=0`
Making terms in the bracket perfect squares
`3[(x-3/2)^2-9/4]-2[(y-1)^2-1]-8=0`
Taking the constant terms out of bracket
`3(x-3/2)^2-27/4-2(y-1)^2-2-8=0`
`=>3(x-3/2)^2-2(y-1)^2-67/4=0`
Dividing both sides with `67/4` and taking `1` to RHS
`=>(3(x-3/2)^2)/(67/4)-(2(y-1)^2)/(67/4)=1`
`=>((x-3/2)^2)/(1/3xx67/4)-(y-1)^2/(1/2xx67/4)=1`
`=>((x-3/2)^2)/(sqrt(67/12))^2-(y-1)^2/(sqrt(67/8))^2=1`

Comparing with general expression of hyperbola

`(x-h)^2/a^2-(y-k)^2/b^2=1`
We get
`a=sqrt(67/12)`, `b=sqrt(67/8)`
from values of `h and k` centre `(3/2,1)`
asymptotic lines as:

`y - k = +- (b/a)(x - h)`
And remaining items: foci and vertices
Cheers.