Question 1152643
I believe the equation was meant to be {{{Ux^2 + y^2 - 2Ux = 0}}} .
 
FACTS ABOUT HYPERBOLAS:
The equation for a hyperbola centered at (h,k) can be written as
{{{(x-h)^2/a^2-(y-k)^2/b^2=" "+- 1}}} for some pair (a,b) of positive numbers.
For {{{(x-h)^2/a^2-(y-k)^2/b^2=-1}}} <--> {{{(y-k)^2/b^2-(x-h)^2/a^2=1}}}  <--> {{{(y-k)^2/b^2=(x-h)^2/a^2+1}}} , you see that it must be {{{(y-k)^2/b^2>=1}}} --> {{{abs(y-k)>=b}}} , so the graph will have an upper branch and a lower branch, like this: 
{{{drawing(300,300,1,19,1,19,circle(10,15,0.1),circle(10,5,0.1),
locate(9.8,16,F),locate(9.8,4.7,"F'"),locate(12,9.95,a),
locate(10.2,12,b),locate(10.2,9,b),locate(12,11.7,c),
rectangle(6,7,10,13),rectangle(6,7,14,10),rectangle(6,7,14,13),
arc(10,10,10,10,-150,-32),arc(10,10,10,10,32,150),
graph(300,300,1,19,1,19,2.5+0.75x,17.5-0.75x,10+sqrt(9+0.5625(x-10)^2),
10-sqrt(9+0.5625(x-10)^2))
)}}} The red and green lines are the asymptotes, with slopes {{{b/a}}} and {{{-b/a}}} .
In the other hand, for {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} <--> {{{(x-h)^2/a^2+(y-k)^2/b^2+1}}} , so it must be {{{(x-h)^2/a^2>=1}}} <--> {{{abs(x-h)>=a}}} , so thee is a left branch and a tight branch to the graph, that looks like this: {{{drawing(300,300,1,19,1,19,
rectangle(6,7,10,13),rectangle(6,7,14,10),rectangle(6,7,14,13),
graph(300,300,1,19,1,19,2.5+0.75x,17.5-0.75x,10+sqrt(0.5625(x-10)^2-9),
10-sqrt(0.5625(x-10)^2-9))
)}}}
 
SOLVING THE PROBLEM:
Adding {{{U}}} to both sides of {{{Ux^2 + y^2 - 2Ux = 0}}} , we get
{{{Ux^2-2Ux+U+y^2=U}}} --> {{{U(x^2-2x+1)+y^2=U}}} --> {{{U(x-1)^2+y^2=U}}} --> {{{highlight((x-1)^2+y^2/U=1)}}}
That equation represents a hyperbola for any negative value of {{{U}}} .
If allowed to choose a value for {{{U}}} to calculate distances, I would choose {{{U=-1}}} .
That makes the equation {{{(x-1)^2-y^2=1}}} , matching the general equation {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} with {{{system(h=1,k=0,a=1,b=1)}}}.
The equation represents a hyperbola with center (1,0),
vertices {{{V(0,0)}}} and {{{W(2,0)}}} , asymptotes {{{y=" "+- (x-1)}}} , focal distance {{{c=sqrt(a^2+b^2)=sqrt(2)}}} , and foci {{{F}}} and {{{"F'"}}} at {{{"("}}}{{{1 +- sqrt(2)}}}{{{", 0)"}}} .
The distance between the foci is {{{2sqrt(2)}}} .
{{{drawing(400,400,-2,4,-3,3,rectangle(0,-1,2,1),
graph(400,400,-2,4,-3,3,x-1,1-x,sqrt((x-1)^2-1),-sqrt((x-1)^2-1)),
circle(0,0,0.05),circle(2,0,0.05),locate(0.05,0.25,V),locate(1.85,0.25,W),
circle(1+sqrt(2),0,0.05),circle(1-sqrt(2),0,0.05),
locate(2,2.2,red(y=x-1)),locate(2,-2,green(y=-x+1)),
locate(-0.5,-0.07,F),locate(2.35,-0.07,"F'")
)}}}
If a generic {{{U}}} is expected, we can say {{{U=-b^2}}} for any {{{b>0}}}
Then we have {{{(x-1)^2-y^2/b^2=1}}} .
That equation represents a hyperbola with center (1,0),
vertices {{{V(0,0)}}} and {{{W(2,0)}}} , asymptotes {{{y=" "+- b(x-1)}}} , focal distance {{{c=sqrt(1+b^2)}}} , and foci at {{{"("}}}{{{1 +- sqrt(1+b^2)}}}{{{", 0)"}}} .
The distance between the foci is {{{2sqrt(1+b^2)}}} .