calculus

find the equation of the tangent line to the curve at the given point by eliminating the parameter:
x=1+sqrt(t) y=e^(t^2) parameter: (2,e)

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  1. x = 1+√t
    t = (x-1)^2
    y = e^(x-1)^4
    y' = e^(x-1)^4 * 4(x-1)^3
    y'(2) = 4e

    Check. Using the parameters, at t=1,
    dy/dt = 2t e^t^2 = 2e
    dx/dt = 1/(2√t) = 1/2
    dy/dx = 2e/(1/2) = 4e

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    oobleck

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