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    Hello for this equation,

    

x^{3} + y^{3} - 3axy =0

    There is a slanted asymptote. I know the above can be written as a parametric form, but just considering the above equation how can you best work out the equation of the asymptote?

    The gradient is simple to find by considering dy/dx, etc..this is equal to -1, but puzzled about finding constant in y=-x + C ...

    Thanks everyone!
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    (Original post by xixixaxa)
    Hello for this equation,

    

x^{3} + y^{3} - 3axy =0

    There is a slanted asymptote. I know the above can be written as a parametric form, but just considering the above equation how can you best work out the equation of the asymptote?

    The gradient is simple to find by considering dy/dx, etc..this is equal to -1, but puzzled about finding constant in y=-x + C ...

    Thanks everyone!
    You have done nearly all the work to find the oblique asymptote. Solve the two equations simultaneously and you will form a quadratic in x. Remember, the asymptote meets the curve at an infinitely distant point. Look out for division by zero.
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    Here's another way to look at it:

    x^3+y^3-3axy=0 can be written as x+y=\frac{3a}{\frac{x}{y}+\frac{  y}{x}-1}.

    Note that \frac{x}{y} and \frac{y}{x} both tend towards {-1} as x tends towards \pm \infty and this should enable you to deduce the equation of the asymptote.
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    (Original post by Mr M)
    You have done nearly all the work to find the oblique asymptote. Solve the two equations simultaneously and you will form a quadratic in x. Remember, the asymptote meets the curve at an infinitely distant point. Look out for division by zero.

    Hey, yes I understand that putting y=mx+c into the first equation will help, and read some sort of solution. The coefficients of x^3 nd x^2 were equated to zero, I don't understand why. I looked at solving a cubic equation, but don't really see why that happened. I know doing that will yield the solution, but can't see it right now.
    I like the other solution a lot better, but please can you explain your first one a little bit more?
    Thanks for you help by the way, really appreciate it!
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    (Original post by xixixaxa)
    The coefficients of x^3 nd x^2 were equated to zero, I don't understand why.
    Can you post the solution you're working from? We can't do anything unless we know what you're looking at.
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    (Original post by xixixaxa)
    Hey, yes I understand that putting y=mx+c into the first equation will help, and read some sort of solution. The coefficients of x^3 nd x^2 were equated to zero, I don't understand why. I looked at solving a cubic equation, but don't really see why that happened. I know doing that will yield the solution, but can't see it right now.
    I like the other solution a lot better, but please can you explain your first one a little bit more?
    Thanks for you help by the way, really appreciate it!
    I'm not totally certain what you are asking but this book offers a very similar method that does not require you to differentiate at all.

    http://books.google.com/books?id=9aA...age&q=&f=false
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    (Original post by generalebriety)
    Can you post the solution you're working from? We can't do anything unless we know what you're looking at.
    (Original post by Mr M)
    I'm not totally certain what you are asking but this book offers a very similar method that does not require you to differentiate at all.

    http://books.google.com/books?id=9aA...age&q=&f=false
    Yes it was exactly the link above. I don't understand why it was equated to zero. ''Roots of this equation are infinitely great" ...

    Thanks guys, much appreciated.
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    (Original post by xixixaxa)
    Yes it was exactly the link above. I don't understand why it was equated to zero. ''Roots of this equation are infinitely great" ...

    Thanks guys, much appreciated.
    Ok I'll try and lead you through the simpler version. I will try to dumb down (no offence intended!) the explanation by avoiding any discussion of limits as I don't know how much mathematics you have.

    You said you already knew m = -1 from differentiation so you have these simultaneous equations:

    y^3 + x^3 - 3axy = 0 ----- (1)

    y = c - x -----(2)

    Substitute (2) into (1) to eliminate y:

    (c-x)^3 +x ^3 - 3ax(c-x)=0

    Expand and collect like terms to form a quadratic equation:

    3(a+c)x^2 -3c (a+c)x +c^3 = 0

    Now think about the denominator of the quadratic formula. In this case it would be 6(a + c).

    You are trying to create a situation where the solutions to this equation are undefined because the asymptote does not ever quite become a tangent to the curve. c=-a will give you division by zero and this is the answer you are looking for.
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    (Original post by Mr M)
    Ok I'll try and lead you through the simpler version. I will try to dumb down (no offence intended!) the explanation by avoiding any discussion of limits as I don't know how much mathematics you have.

    You said you already knew m = -1 from differentiation so you have these simultaneous equations:

    y^3 + x^3 - 3axy = 0 ----- (1)

    y = c - x -----(2)

    Substitute (2) into (1) to eliminate y:

    (c-x)^3 +x ^3 - 3ax(c-x)=0

    Expand and collect like terms to form a quadratic equation:

    3(a+c)x^2 -3c (a+c)x +c^3 = 0

    Now think about the denominator of the quadratic formula. In this case it would be 6(a + c).

    You are trying to create a situation where the solutions to this equation are undefined because the asymptote does not ever quite become a tangent to the curve. c=-a will give you division by zero and this is the answer you are looking for.
    Ah of course!! Sorry...When I saw the solution of the link posted, I tried to equate it to the method for finding the roots of a cubic equation assuming that I used y=mx+c instead... Coefficient of  x^3 is equated to 0 to produce a quadratic, and then coefficient of x^2 is equated to 0 for the reason you mentioned...Is that right?

    I should know enough about limits yes...is it possible for you to briefly tell me what you were going to say about them? I want to make sure I understand this in different ways as best I can...thank you very much
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    (Original post by xixixaxa)
    Ah of course!! Sorry...When I saw the solution of the link posted, I tried to equate it to the method for finding the roots of a cubic equation assuming that I used y=mx+c instead... Coefficient of  x^3 is equated to 0 to produce a quadratic, and then coefficient of x^2 is equated to 0 for the reason you mentioned...Is that right?

    I should know enough about limits yes...is it possible for you to briefly tell me what you were going to say about them? I want to make sure I understand this in different ways as best I can...thank you very much
    I think this subject is a bit exhausted now really.

    The wikipedia entry on Asymptotes is good.

    http://en.wikipedia.org/wiki/Asymptote
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    (Original post by xixixaxa)
    Coefficient of  x^3 is equated to 0 to produce a quadratic
    No, all the x^3 terms just happen to cancel out.

    (Original post by xixixaxa)
    coefficient of x^2 is equated to 0 for the reason you mentioned...
    Correct, but not specifically because it's the coefficient of x^2, rather because it makes the solutions of that quadratic equation 'infinite'.

    No one is "equating" any coefficients "to 0". Try and follow the logic rather than looking for some sort of secret method!
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    (Original post by generalebriety)
    No, all the x^3 terms just happen to cancel out.


    Correct, but not specifically because it's the coefficient of x^2, rather because it makes the solutions of that quadratic equation 'infinite'.

    No one is "equating" any coefficients "to 0". Try and follow the logic rather than looking for some sort of secret method!
    I was considering just y=mx+c...it is more cumbersome to work out the gradient. Without knowing m or c, you'd get a cubic - so it's equated to 0 to form a quadratic and then from that you would make the solutions of the quadratic equation 'infinite'?

    Thanks
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    (Original post by xixixaxa)
    Without knowing m or c, you'd get a cubic - so it's equated to 0 to form a quadratic
    Yet again, no. Without knowing m or c you have one equation in two unknowns and can't solve it. We can't just remove terms because they make it hard to solve.

    (Original post by xixixaxa)
    from that you would make the solutions of the quadratic equation 'infinite'?
    Yep.
 
 
 
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