Question 726334
convert the following equation to the standard form of a conic section: 
x^2-y^2-6x+13=0 
complete the square:
x^2-6x-y^2+13=0
(x^2-6x+9)-y^2=-13+9
(x-3)^2-y^2=-4
divide by -4
{{{y^2/4-(x-3)^2/4=1}}}
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=(x,y) coordinates of the center.
For given hyperbola:
center: (3,0)
a^2=b^2=4