# Thread: Cosine/hyperbola qiestion

1. ## Cosine/hyperbola qiestion

Can someone please explain this to me?

Any help would be greatly appreciated!

2. (This is more of a Precalculus question, I think.)

$\frac{x^2}{25} - \frac{y^2}{144} = 1$

Comparing this to the basic form:
$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

The center is at (0, 0), a = 5, and b = 12. Use the Pythagorean relation
$c^2 = a^2 + b^2$ to solve for c:
\begin{aligned}
c^2 &= a^2 + b^2 \\
c^2 &= 5^2 + 12^2 \\
c^2 &= 169 \\
c &= 13
\end{aligned}

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:
$y = A\cos[B(x - C)]$
(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 $\frac{2\pi}{|b|}$, and can be found by finding the distance from one maximum to the next, which in our case is 26.
\begin{aligned}
\frac{2\pi}{|B|} &= 26 \\
2\pi &= 26|B| \\
|B| &= \frac{\pi}{13} \\
B &= \pm \frac{\pi}{13}
\end{aligned}

For our purpose we'll just use the positive value, $B = \frac{\pi}{13}$.

In $y = \cos x$, one of the maximum is at (0, 1).
In $y = A\cos x$, 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 $C = -13$:
$y = A\cos\left[\frac{\pi}{13}(x - (-13))\right]$
$y = A\cos\left[\frac{\pi}{13}(x + 13)\right]$

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).

01