(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.

1 Answer
Jun 15, 2016

Answer:

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.