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Thread: Asymptotes of a Hyperbola

  1. #1
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    Exclamation Asymptotes of a Hyperbola

    How do you find the equations of the asymptotes of this hyperbola?

    -x^2 + y^2 + 6x - 16y - 9 = 0
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  2. #2
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    Quote Originally Posted by jumpman23 View Post
    How do you find the equations of the asymptotes of this hyperbola?

    -x^2 + y^2 + 6x - 16y - 9 = 0
    1. By completing the square you get:

    -(x^2-6x+9) + (y^2-16y+64)=9-9+64~\implies~-\dfrac{(x-3)^2}{64}+\dfrac{(y-8)^2}{64}=1

    2. The center of the hyperbola is C(3, 8).
    The semi-axes are a = 8 and b = 8

    3. Therefore the asymptotes have the slopes m_1 = 1~\vee~m_2=-1

    4. Use the point-slope-formual to get the equations of the asymptotes:

    a_1: y= x+5

    a_2: y = -x +11
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  3. #3
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    Here's another way of arguing. The "asymptotes" of a hyperbola are the lines the hyperbola approaches for very very large values of x and y.

    The equation of the hyperbola is -\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 1 as Earboth said. For very very large x and y those squares will each be far larger! Large enough that that "1" on the right side is negligible compared to them: For very very large x and y, that equation is very very close to -\frac{(x- 3)^2}{64}+ \frac{(y- 8)^2}{64}= 0 from which we get \frac{(x- 3)^2}{64}= \frac{(y- 8)^2}{64} and then  x- 3= \pm (y- 8). Taking x- 3= y- 8, y= x+ 5. Taking x- 3= -(y- 8), x- 3= -y+ 8, y= -x+ 11, the answer that Earboth gave.
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