Question 1039199
The center is the intersection of the asymptotes.
That is the point that is a solution to {{{system(y=(9/8)x+1,y=-(9/8)x-9)}}} .
You can solve that system of equations by substitution.
For example, you could start by substituting the expression {{{(9/8)x+1}}} for {{{y}}} in {{{y=-(9/8)x-9}}} .
{{{system(y=(9/8)x+1,y=-(9/8)x-9)}}} --> {{{system(y=(9/8)x+1,(9/8)x+1=-(9/8)x-9)}}} --> {{{system(y=(9/8)x+1,(9/8)x+(9/8)x=-9-1)}}} --> {{{system(y=(9/8)x+1,(9/4)x=-10)}}} --> {{{system(y=(9/8)x+1,x=-10(4/9))}}} --> {{{system(y=(9/8)x+1,x=-40/9)}}} ,
and then, you could substitute the value found for one variable into one of the equations, to find the value of the other variable:
{{{system(y=(9/8)x+1,x=-40/9)}}} --> {{{system(y=(9/8)(-40/9)+1,x=-40/9)}}} --> {{{system(y=(9/8)(-40/9)+1,x=-40/9)}}} --> {{{system(y=-5+1,x=-40/9)}}} --> {{{highlight(system(y=-4,x=-40/9))}}}
 
The asymptotes slopes give you the {{{b/a=9/8}}} ratio of conjugate to transverse axes.
The transverse axis is the segment joining the vertices of the hyperbola.
If the length is {{{14}}} , the distance from the center to each vertex must be
{{{a=14/2=7}}} .
From there you can find {{{b}}} , {{{c=sqrt(b^2+a^2)}}} , and the eccentricity: {{{c/a}}} .