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    Find an equation of the line which is a tangent to both the parabola with equation y^2 = 4ax and the parabola with equation x^2 = 4ay.
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    Well, those are inverse functions, aren't they? So they must be reflected in y=x...
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    (Original post by henryt)
    Well, those are inverse functions, aren't they? So they must be reflected in y=x...
    thanks for that. :rolleyes:
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    Let's consider the general point (at^2, 2at) on the first parabola. The equation of the tangent at that point is:
    y = (1/t)x + at

    Now, let's see where this tangent intersects the other parabola:
    x^2 = 4a((1/t)x + at) = (4a/t)x + 4a^2t
    => x^2 - (4a/t)x - 4a^2t = 0

    We want them to intersect only once, so:
    (4a/t)^2 - 4 * 1 * (-4a^2t) = 0
    => t^3 = -1
    => t = -1

    So, the line is:
    y = -(x + a)
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    (Original post by Popcorn)
    thanks for that. :rolleyes:
    Well, sorry if I haven't actually done FP2 yet, and don't know what I'm talking about. Sheesh.
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    Just a side note: they're not really inverse functions (of each other).
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    (Original post by dvs)
    Let's consider the general point (at^2, 2at) on the first parabola. The equation of the tangent at that point is:
    y = (1/t)x + at

    Now, let's see where this tangent intersects the other parabola:
    x^2 = 4a((1/t)x + at) = (4a/t)x + 4a^2t
    => x^2 - (4a/t)x - 4a^2t = 0

    We want them to intersect only once, so:
    (4a/t)^2 - 4 * 1 * (-4a^2t) = 0
    => t^3 = -1
    => t = -1

    So, the line is:
    y = -(x + a)
    Sorry a couple of years late but I don't understand what you did so that it intersects only once, can you explain this to me. thank you
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    anyone?
 
 
 
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Updated: June 5, 2009
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