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    I don't understand how the answer to this is y = +/- (1/3)x

    The question is to find the equations of the asymptotes to:
    ((x^2)/9) - y^2 = 1.

    Thanks, and sorry for my stupidity, I can find them when there are denominators etc, but I just can't figure out this one.
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    (Original post by Konnichiwa)
    I don't understand how the answer to this is y = +/- (1/3)x

    The question is to find the equations of the asymptotes to:
    ((x^2)/9) - y^2 = 1.

    Thanks, and sorry for my stupidity, I can find them when there are denominators etc, but I just can't figure out this one.
    Are you sure it's not just a formula you have to remember?

    \left( \dfrac{x}{3} \right)^2 - \left(\dfrac{y}{1} \right)^2=1

    EDIT: Oh, did you mean "there are denominators" as in rational functions? If so, I guess rewriting it doesn't help, haha.
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    (Original post by aznkid66)
    Are you sure it's not just a formula you have to remember?

    \left( \dfrac{x}{3} \right)^2 - \left(\dfrac{y}{1} \right)^2=1

    EDIT: Oh, did you mean "there are denominators" as in rational functions? If so, I guess rewriting it doesn't help, haha.
    That doesn't look fp1 to me..
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    I meant denominators with variables in, I sorta rushed my post, sorry. :P

    And it is from FP1. Jan12 Q7.

    Edit: I just don't get how to find the asymptote y = +/- a third x.
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    Well, do you know what a hyperbola looks like?
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    Yes.
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    Well, for a hyperbola:

    \left( \dfrac{x-h}{a} \right) ^2 - \left( \dfrac{y-k}{b} \right) ^2 = 1

    The asymptotes are:

    y=\pm \dfrac{b}{a} (x-h) + k

    You can get this from the equation of a hyperbola in the problem you showed. There are no points on the hyperbola above or below the asymptotes. In this case, "above or below the asymptotes" means where "y is too positive" or "y is too negative". So what are the bounds for y in terms of x? Well, if |y|≥x/3, then the difference in the LHS of the equation would be less than or equal to 0 for all x values. Because this is the case, we know that when |y|≥x/3, the LHS will never equal to RHS, and there will never be points on the hyperbola where |y|≥x/3. On the other hand, if |y|<x/3, then there will be an x value such that the first term is exactly 1 unit greater than the second term.
    Yeah, in hindsight, I actually have no clue where those equations come from ^^;
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    Thanks, that was a clear explanation.

    Is this in the formula book or will I have to learn it by heart? I can't find the latest official GCE mathematics formula booklet on the AQA website.
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    Oh!

    ((x^2)/9) - y^2 = 1

    Means that:

    (x/3+y)(x/3-y)=1

    Which means that y≠±x/3...

    Okay, I still don't know.
    Are you expected to know calculus, by FP1, yet? You could just show that the gradient of the upper-right and lower-left branch approaches 1/3 and the gradient of the lower-right and upper-left branch approaches -1/3.
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    If this is the formula book you're using, then yeah, it's on there.

    Attachment 207696
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    (Original post by aznkid66)
    If this is the formula book you're using, then yeah, it's on there.

    Attachment 207696
    What page is it on? I can't find it at all.
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    Well, it should be in the "conics" section. Which board are you doing? For edexcel, it's not in FP1 (but it's in FP3), so I don't think you'd be tested on it.
 
 
 
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