SOLUTION: Find the equation of the hyperbola (in standard and general forms) which satisfies the conditions given. Foci at (2-√13,3) and (2+√13,3), vertices at (5,3) and (-1,3)

Instead of doing your problem for you, I'm doing one which is EXACTLY
STEP-BY-STEP like yours.  All you have to do is use it as a model and
follow it step-by-step.  Here's the problem that I'll solve for you: 

We plot the center and foci to see whether the hyperbola is like 
this ) ( or like this: 



The hyperbola is like this ") (",so its equation is of this form:



where (h,k) = the center = (4,5), 
and "a" = the distance from the center to the vertex, which is the
distance from (4,5) to (9,5) which is 9-4=5

so we have this much of the 
equation:



The distance from the center to either focus is the value c,
and that is the distance from (4,5) to (4+√41,5), which is 
(4+√41)-(4) = √41
and the Pythagorean relationship for all hyperbolas is

c2=a2+b2.  






  <---that's the standard form.

That's the standard form equation. To draw the graph, we make a defining
rectangle which is 2a=10 units wide and 2b=8 units high, with the center (4,5)
as the center of the defining rectangle, and draw its extended diagonals,
which are the asymptotes.



Now we can sketch in the hyperbola:



To get the general form we start with the standard form:





We multiply through by the LCD = (25)(16) = 400















  <--that's the general form.

Now go do yours the exact same way, step by step.

You weren't asked to draw the graph, but it helps.

Edwin