Question 426192
What is the general conic form equation of the hyperbola?
(y+1)^2/49-(x+8)^2/196=1
..
Standard forms of hyperbolas:
If y^2 term comes first: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center
If x^2 term comes first: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center, as with the first form. 

Given equation is of the first form with center at (-8,-1) and vertical transverse axis, that is, the hyperbola opens up and down.
a^2=49
a=7 (distance from center to vertices on the transverse axis.
Length of transverse axis=2a=14
b^2=196
b=14
Length of conjugate axis=2b=28
c^2=a^2+b^2=245
c=sqrt(245)=15.65..(distance from center to foci on the transverse axis)
Equation of asymptotes:(use y=mx+b form, with m=+-a/b=+-7/14=+-1/2, and line going thru center (-8,-1)
Two formulas:
y=x/2+3
y=-x/2-5
see the graph below:

..
y=+-(49+49(x+8)^2/196)^.5-1

{{{ graph( 400, 400, -30, 30, -30, 30, (49+49(x+8)^2/196)^.5-1,-(49+49(x+8)^2/196)^.5-1,x/2+3,-x/2-5) }}}