Question 1043331
{{{25x^2-39y^2+150x+390y=-225}}}
{{{25x^2+150x-39y^2+390y=-225}}}
{{{25(x^2+6x)-39(y^2-10y)=-225}}}
{{{25(x^2+6x+9)-39(y^2-10y+25)=-225+25*9-39*25}}}
{{{25(x^2+6x+9)-39(y^2-10y+25)=-225+225-975}}}
{{{25(x+3)^2-39(y-5)^2=-975}}}
{{{39(y-5)^2-25(x+3)^2=975}}}
{{{39(y-5)^2/975-25(x+3)^2/975=975/975}}}
{{{highlight((y-5)^2/25-(x+3)^2/39=1)}}}
The standard equation above tells us that
the center is at (-3,5) ;
the major (or transverse) axis is {{{x=-3}}};
the focal distance is {{{c=sqrt(25+39)=sqrt(64)=8}}} ; 
the asymptotes have slopes such that {{{slope^2=25/39}}} , and
the y-coordinates of the vertices can be found from
{{{(y-5)^2/25=1}}} .
The asymptotes pass through center {{{"(-3 , 5 )"}}} ,
and have slopes such as
{{{slope^2=25/39}}}-->{{{system(slope=5sqrt(39)/39,"or",slope=-5sqrt(39)/39)}}} ,
the equations of the asymptotes (in point-slope form) are
{{{highlight(y-5=5sqrt(39)(x+3)/39)}}} and {{{highlight(y-5=-5sqrt(39)(x+3)/39)}}} .
Those equations can also be written as
{{{highlight(y=-5sqrt(39)x/39+(65-15sqrt(39))/13)}}} and
{{{highlight(y=5sqrt(39)x/39+(65+15sqrt(39))/13)}}} in slope-intercept form.
Since the foci are on the major axis,
at a distance {{{c=8}}} from the center,
their y- coordinates are {{{y=5 +- 8}}} , or {{{system(y=5+8=13,"or",y=5-8=-3)}}} .
So, one focus is at {{{highlight("( -3 , 13 )")}}} ,
and the tother focus is at {{{highlight("( -3 , -3 )")}}}
As for the vertices, also on the major axis,
{{{(y-5)^2/25=1}}}-->{{{(y-5)^2=25}}}-->{{{y=5 +- sqrt(25)= 5 +- 5}}}---> {{{y=10,"or",y=0)}}} .
So the vertices are at {{{highlight("( -3 , 10 )")}}} and {{{highlight("( -3 , 0 )")}}} .
The hyperbola with major axis and foci looks like this:
{{{drawing(300,300,-30,30,-5,15,
graph(300,300,-30,30,-5,15,5+sqrt(1+(x+3)^2/39),-6,5-sqrt(1+(x+3)^2/39)),
green(line(-3,-6,-3,20)),
circle(-3,-3,0.5),circle(-3,13,0.5)
)}}}