# Cosine/hyperbola qiestion

• Jul 25th 2009, 12:51 PM
s3a
Cosine/hyperbola qiestion
Can someone please explain this to me?

Any help would be greatly appreciated!
• Jul 25th 2009, 03:36 PM
yeongil
(This is more of a Precalculus question, I think.)

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

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

The center is at (0, 0), a = 5, and b = 12. Use the Pythagorean relation
$\displaystyle c^2 = a^2 + b^2$ to solve for c:
\displaystyle \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:
$\displaystyle y = A\cos[B(x - C)]$
(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 $\displaystyle \frac{2\pi}{|b|}$, and can be found by finding the distance from one maximum to the next, which in our case is 26.
\displaystyle \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, $\displaystyle B = \frac{\pi}{13}$.

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

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).

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