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Math Help - conics

  1. #1
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    conics

    Compute the points of the intersection between the circle x^2+y^2=1 and hyperbola xy=1?

    I tried to put 1/y in place of x but I couldn't solve.
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  2. #2
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    Re: conics

    Quote Originally Posted by kastamonu View Post
    Compute the points of the intersection between the circle x^2+y^2=1 and hyperbola xy=1?

    I tried to put 1/y in place of x but I couldn't solve.
    You have x^2+y^2=1 and xy=1.

    We may assume x,y\not=0 since they don't lie on the hyperbola anyways.

    Taking x=1/y as you have tried,

    \frac{1}{y^2}+y^2=1

    Multiplying by y^2 and rearranging gives

    y^4-y^2+1=0

    which has no real solutions. So there is no intersection. In fact, this is obvious if you graphed the two functions.
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    Re: conics

    There is an answer. By discriminant we can find the roots. t^2=x^4 and we can find the roots.
    \Delta = \,b^2-4ac.
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    Re: conics

    Yes, but the discriminant is negative, so all the roots are non-real numbers in the complex plane. That is to say there is no real solution. As I have suggested earlier, just graphing these two functions makes it clear that they do not intersect on the xy plane.
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    Re: conics

    conics-graph.png
    Thanks from Gusbob
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  6. #6
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    Re: conics

    It is positive.
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    Re: conics

    good drawing.
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    Re: conics

    Quote Originally Posted by Gusbob View Post
    Yes, but the discriminant is negative, so all the roots are non-real numbers in the complex plane. That is to say there is no real solution. As I have suggested earlier, just graphing these two functions makes it clear that they do not intersect on the xy plane.
    It is positive.
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    Re: conics

    The equation reduces to t^2-t+1=0 when we put t = y^2.
    Now discriminant of equation is b^2-4ac= (1)^2- 4*1*1= 1 - 4 = -3 NEGATIVE
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  10. #10
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    Re: conics

    ıt was x^2+y^2=4.I made a mistake.
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    Re: conics

    Kastamonu ...........

    no further comments

    conics-sol002.png
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    Re: conics

    In that case the equation reduces to t^2+4t-4=0; Now the discriminant = 4^2-4(1)(-4)= 32 so we will her discriminant = 4sqrt2
    take the positive root because t = x^2 and proceed to get the required points.
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  13. #13
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    Re: conics

    many thanks.
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