calculus

the tangent line to the curve y=x^2 at the point (a,a^2) passes through the point (2,1) find all possible values of a.
Thank you so much!!

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  1. dy/dx = 2x
    at (a,a^2) , dy/dx = 2a

    equation of tangent:
    y - a^2 = 2a(x-a)
    at 2,1)
    1 - a^2 = 2a(2-a)
    1 - a^2 = 4a - 2a^2
    a^2 - 4a + 1 = 0
    let's complete the square:
    a^2 - 4a + 4 = -1 + 4
    (a-2)^2 = 3
    a-2 = ± √3
    a = 2 ± √3

    or

    let point of contact be (a,a^2) , let P(2,1) be the outside point.
    dy/dx = 2a
    so at (a,a^2) slope = 2a
    grade 9 way: slope = (a^2-1)/(a-2)

    so (a^2 - 1)/(a-2) = 2a
    2a^2 - 4a = a^2 - 1
    a^2 - 4a + 1 = 0
    same as before

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