Question 960974
Find the vertices and foci of the hyperbola.
9x2 − y2 − 36x − 2y + 26 = 0
***
Rewrite:
9x^2-36x-y^2-2y=-26 
complete the square:
9(x^2-4x+4)-(y^2+2y+1)=-26+36-1
9(x-2)^2-(y+1)^2=9
{{{(x-2)^2-(y+1)^2/9=1}}}
hyperbola has a horizontal transverse axis.
Its standard form of equation: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=coordinates of center
center: (2, -1)
a^2=1
a=1
b^2=9
b=3
vertices: (2±a,-1)=(2±1,-1)=(1, -1) and (3,-1)
c^2=a^2+b^2=1-9=10
c=√10≈3.16
foci: (2±c,-1)=(2±3.16,-1)=(1.16, -1) and (5.16,-1)