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March 29, 2017

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There are two points A=(2,0) and B=(4,0) and a moving point P=(t,t) on the xy-plane. If the minimum value of line AP+ line BP is b, what is the value of b^2?

  • Math - ,

    AP = √((t-2)^2 + t^2)
    BP = √((t-4)^2 + t^2)

    b^2 = (AP+BP)^2
    = AP^2 + 2 AP*BP + BP^2
    = ((t-2)^2 + t^2)+((t-4)^2 + t^2) + 2√(((t-2)^2 + t^2)*((t-4)^2 + t^2))
    = 4(t^2 - 3t + 5 + √(t^4-6t^3+18t^2-24t+16))

    min b^2 = 20 when t = 4/3

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