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Thread: Cosine/hyperbola qiestion

  1. #1
    s3a
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    Cosine/hyperbola qiestion

    Can someone please explain this to me?

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    (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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