Question 1152644
Find the value/s of U so that the graph of the equation of 
Ux^2 + y^2 − 2Ux = 0 is a hyperbola.

Afterwards, identify the distance/s between the foci.
<pre>
That will be the equation of a hyperbola if you pick any negative number
for U.  The easiest negative number to pick for U is -1:

{{{Ux^2 + y^2 − 2Ux = 0}}}

{{{(-1)x^2 + y^2 − 2(-1)x = 0}}}

{{{-x^2 + y^2 + 2x = 0}}}

{{{y^2 - x^2 + 2x = 0}}}

Write y² as (y-0)² and factor out a negative from the two terms in x:

{{{(y-0)^2 - (x^2-2x)=0}}}

Complete the square by adding +1 inside the parentheses, which
amounts to adding -1 to the left side, so add -1 to the right
side also:

{{{(y-0)^2 - (x^2-2x+1)=-1}}}

We write the second parentheses as the square of a binomial:

{{{(y-0)^2 - (x-1)^2=-1}}}

Since the right side is negative, we multiply through by -1

{{{-(y-0)^2 + (x-1)^2=+1}}}

We reverse the terms on the left:

{{{(x-1)^2 - (y-0)^2=1}}}

We put 1's under the two terms on the left:

{{{(x-1)^2/1^"" - (y-0)^2/1^""=1}}}


{{{drawing(400,800/3,-2,4,-2,2, graph(400,800/3,-2,4,-2,2,-sqrt(x^2-2x)),

graph(400,800/3,-2,4,-2,2,sqrt(x^2-2x)) )}}}

We compare 

{{{(x-1)^2/1^"" - (y-0)^2/1^""=1}}} to

{{{(x-h)^2/a^2 - (y-k)^2/b^2=1}}}

The center is (h,k) = (1,0), a=1, b=1,

We calculate c, the distance fro center to foci from 

{{{c^2=a^2+b^2}}}
{{{c^2=1^2+1^2}}}
{{{c^2=1+1}}}
{{{c^2=2}}}
{{{c=sqrt(2)}}}

So the foci are {{{(matrix(1,3,1+sqrt(2),",",0))}}} and {{{(matrix(1,3,1-sqrt(2),",",0))}}},

so the distance between the foci is

{{{2sqrt(2)}}}

If you chose any other negative number for U the answer would be different.

Edwin</pre>