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    The curve is given by the implicit equation xy^2 = 8, x>0. The line is the tangent at point P with coordinates (2,2) this line meets the curve again at Q.

    (From various parts of the question
    dy/dx = -y/2x
    tangent at P dy/dx =1/2
    co-ordinates of Q (8,-1)
    parametric eq: x = 2t^2 , y=2/t
    dy/dx = -1/(2t^3) ...in terms of t

    Question: Prove if k > 3 sqrcube(2) the line x +y = k cuts the curve in 3 places.

    now they said the answer k > sqr^3(2) + 2sqr^3(2)
    but how does this show it cuts in 3 places exactly? Is it bcos it's sqr^3? But surely you wouldn't have to work all that out if that's all it is. I don't get it Thank you..
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    See if you can work it out from the attached graph.
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    (Original post by Jonny W)
    See if you can work it out from the attached graph.
    The only line I can see that cuts 3 times is the second to top one? Um but 2nd intersect of the line, one of the x + y = 2.4 which is under k = 3.7.. ? hmm
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    I mean I see how they got the answer:
    gradient of x + y = k is -1
    so dy/dx = -1 = -1/(2t^3)
    t = 1/[sqr^3(2)] so x and y are [sqr^3(2) , 2sqr^3(2)]
    so k > sqr^3(2) + 2sqr^3(2) but I don't see how this last line proves it cuts it at 3 points :confused: I think i'm missing something here
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    (Original post by Maisy)
    The only line I can see that cuts 3 times is the second to top one? Um but 2nd intersect of the line, one of the x + y = 2.4 which is under k = 3.7.. ? hmm
    The top line also cuts the curve three times (once outside the area of the graph). The bottom three lines each cut the curve once.

    The "critical line" is the tangent T to the curve that has the form x + y = k. (This will have equation x + y = about 3.8, since it is slightly below x + y = 4.) Every x + y = constant line above T cuts the curve three times.

    Therefore the problem is to find T.
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    (Original post by Jonny W)
    The top line also cuts the curve three times (once outside the area of the graph). The bottom three lines each cut the curve once.

    The "critical line" is the tangent T to the curve that has the form x + y = k. (This will have equation x + y = about 3.8, since it is slightly below x + y = 4.) Every x + y = k line above T cuts the curve three times.

    Therefore the problem is to find T.
    Thanks I see it now. Well t = 1/[sqr^3(2)] so that gives you x and y [sqr^3(2) , 2sqr^3(2)] which is the points of the tangent I'm guessing so everything above that cuts 3 times.. is that all you can say?
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    (Original post by Sugaray)
    Thanks I see it now. Well t = 1/[sqr^3(2)] so that gives you x and y [sqr^3(2) , 2sqr^3(2)] which is the points of the tangent I'm guessing so everything above that cuts 3 times.. is that all you can say?
    t = 2^(-1/3) gives the point on the curve where T (equation, x + y = k) is tangent. That point is (2*2^(-2/3), 2/2^(-1/3)) = (2^(1/3), 2*2^(1/3)).

    We still haven't worked out k. But since the point (2^(1/3), 2*2^(1/3)) is on T, its components must add up to k. So k = 2^(1/3) + 2*2^(1/3) = 3*2^(1/3) = 3.77976 (5dp).
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