Question 886273
<pre>{{{drawing(400,400,-16,10,-12,14,

rectangle(3,-8,-9,10),graph(400,400,-16,10,-12,14,(2+3sqrt(x^2+6x-27))/2),
green(line(-23,-29,21,37),line(23,-38,-29,40)),
graph(400,400,-16,10,-12,14,(2-3sqrt(x^2+6x-27))/2),

locate(-6.5,1.5,"(-3,1)"),locate(-16,2.5,(matrix(1,3,-3-3sqrt(13),",",1))),
locate(.8,6.8,"(1,7)"),
circle(-13.8166538,1,0.15),circle(-13.8166538,1,0.13),circle(-13.8166538,1,0.11),circle(-13.8166538,1,0.09),circle(-13.8166538,1,0.07),circle(-13.8166538,1,0.05),circle(-13.8166538,1,0.03),circle(-13.8166538,1,0.01),

circle(-3,1,0.15),circle(-3,1,0.13),circle(-3,1,0.11),circle(-3,1,0.09),circle(-3,1,0.07),circle(-3,1,0.05),circle(-3,1,0.03),circle(-3,1,0.01),

circle(1,7,0.15),circle(1,7,0.13),circle(1,7,0.11),circle(1,7,0.09),circle(1,7,0.07),circle(1,7,0.05),circle(1,7,0.03),circle(1,7,0.01)





  )}}}


So we know that the hyperbola has the equation:

{{{(x-h)^2/a^2}}}{{{""-""}}}{{{(y-k)^2/b^2}}} {{{""=""}}} {{{1}}}

The asymptotes intersect at the center, so the center is 

(h,k) = (-3,1).

{{{(x+3)^2/a^2}}}{{{""-""}}}{{{(y-1)^2/b^2}}} {{{""=""}}} {{{1}}}

We only need to know a and b.

The slope of the asymptotes are {{{"" +- b/a}}}

So we find the slope of the asymptote through (-3,1) and (1,7).

We use the slope formula:

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

m = {{{(7-1)/(1-(-3))}}} =  {{{6/(1+3)}}} = {{{6/4}}} = {{{3/2}}}

Therefore  

{{{b/a}}}{{{""=""}}}{{{3/2}}}

All hyperbolas have the Pythagorean relationship {{{c^2=a^2+b^2}}}

We can find the value of " c " because it is the distance from
the focus to the center, so we find the distance 

between the focus {{{(matrix(1,3,-3-3sqrt(13),",",1))}}} and the
center(-3,1)

using the distance formula:

{{{d}}}{{{""=""}}}{{{ sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2)}}} 

{{{c}}}{{{""=""}}}{{{ sqrt( (-3-(-3-3sqrt(13)))^2 + (1-1)^2)}}}

{{{c}}}{{{""=""}}}{{{ sqrt( (-3+3+3sqrt(13))^2 + (0)^2)}}}

{{{c}}}{{{""=""}}}{{{ sqrt( (3sqrt(13))^2) }}}

{{{c}}}{{{""=""}}}{{{sqrt(9*13)}}}

{{{c}}}{{{""=""}}}{{{sqrt(117)}}}

So {{{c^2=117}}}, and since {{{c^2=a^2+b^2}}}

{{{117=a^2+b^2}}}

We solve the system of equations:

{{{system(117=a^2+b^2,b/a=3/2)}}}

Solve the second one for b:

{{{2b=3a}}}
{{{b=3a/2}}}

Substitute in the first equation:

{{{117=a^2+(3a/2)^2}}}
{{{117=a^2+9a^2/4}}}
{{{468=4a^2+9a^2}}}
{{{468=13a^2}}}
{{{468/13=a^2}}}
{{{36=a^2}}}
{{{6=a}}}

Substitute in {{{b=3a/2}}}

{{{b=3(6)/2=9}}}

Therefore the equation of the hyperbola is:

{{{(x+3)^2/6^2}}}{{{""-""}}}{{{(y-1)^2/9^2}}} {{{""=""}}} {{{1}}}

or

{{{(x+3)^2/36}}}{{{""-""}}}{{{(y-1)^2/81}}} {{{""=""}}} {{{1}}}

Edwin</pre>