(This is more of a Precalculus question, I think.)
Comparing this to the basic form:
The center is at (0, 0), a = 5, and b = 12. Use the Pythagorean relation
to solve for c:
The foci are at (c, 0) and (-c, 0), or (13, 0) and (-13, 0).
Now, we look at the form of the cosine function:
(I capitalized A, B & C so that you can see that they are different than the a, b & c I used for the hyperbola equation.)
The period of is , and can be found by finding the distance from one maximum to the next, which in our case is 26.
For our purpose we'll just use the positive value, .
In , one of the maximum is at (0, 1).
In , the corresponding maximum is at (0, A).
In our function, since one of the maxima is at (-13, A) (we don't know what A is yet), our function has a horizontal shift 13 units to the left, or :
At this point, we could solve for A, why bother? There's only one choice that matches our values for B & C, and that is choice (A).