SOLUTION: Find the value/s of U so that the graph of the equation of Ux2 + y^2 − 2Ux = 0 is a hyperbola. Afterwards, identify the distance/s between the foci.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the value/s of U so that the graph of the equation of Ux2 + y^2 − 2Ux = 0 is a hyperbola. Afterwards, identify the distance/s between the foci.      Log On


   

Question 1152643: Find the value/s of U so that the graph of the equation of Ux2 + y^2 − 2Ux = 0 is a hyperbola.
Afterwards, identify the distance/s between the foci.
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
I believe the equation was meant to be Ux%5E2+%2B+y%5E2+-+2Ux+=+0 .

FACTS ABOUT HYPERBOLAS:
The equation for a hyperbola centered at (h,k) can be written as
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=%22+%22%2B-+1 for some pair (a,b) of positive numbers.
For %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=-1 <--> %28y-k%29%5E2%2Fb%5E2-%28x-h%29%5E2%2Fa%5E2=1 <--> %28y-k%29%5E2%2Fb%5E2=%28x-h%29%5E2%2Fa%5E2%2B1 , you see that it must be %28y-k%29%5E2%2Fb%5E2%3E=1 --> abs%28y-k%29%3E=b , so the graph will have an upper branch and a lower branch, like this:
The red and green lines are the asymptotes, with slopes b%2Fa and -b%2Fa .
In the other hand, for %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 <--> %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2%2B1 , so it must be %28x-h%29%5E2%2Fa%5E2%3E=1 <--> abs%28x-h%29%3E=a , so thee is a left branch and a tight branch to the graph, that looks like this:

SOLVING THE PROBLEM:
Adding U to both sides of Ux%5E2+%2B+y%5E2+-+2Ux+=+0 , we get
Ux%5E2-2Ux%2BU%2By%5E2=U --> U%28x%5E2-2x%2B1%29%2By%5E2=U --> U%28x-1%29%5E2%2By%5E2=U --> highlight%28%28x-1%29%5E2%2By%5E2%2FU=1%29
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 %28x-1%29%5E2-y%5E2=1 , matching the general equation %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1 with system%28h=1%2Ck=0%2Ca=1%2Cb=1%29.
The equation represents a hyperbola with center (1,0),
vertices V%280%2C0%29 and W%282%2C0%29 , asymptotes y=%22+%22%2B-+%28x-1%29 , focal distance c=sqrt%28a%5E2%2Bb%5E2%29=sqrt%282%29 , and foci F and %22F%27%22 at %22%28%221+%2B-+sqrt%282%29%22%2C+0%29%22 .
The distance between the foci is 2sqrt%282%29 .

If a generic U is expected, we can say U=-b%5E2 for any b%3E0
Then we have %28x-1%29%5E2-y%5E2%2Fb%5E2=1 .
That equation represents a hyperbola with center (1,0),
vertices V%280%2C0%29 and W%282%2C0%29 , asymptotes y=%22+%22%2B-+b%28x-1%29 , focal distance c=sqrt%281%2Bb%5E2%29 , and foci at %22%28%221+%2B-+sqrt%281%2Bb%5E2%29%22%2C+0%29%22 .
The distance between the foci is 2sqrt%281%2Bb%5E2%29 .