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Math Help - pair of tangents

  1. #1
    Junior Member cinder's Avatar
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    pair of tangents

    Here's the question:



    I don't know how to proceed.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cinder View Post
    Here's the question:



    I don't know how to proceed.
    First thing we do know about these curves, is that both have the point (1,0)

    Let y_1 = x^2 + ax + b
    Let y_2 = cx - x^2

    We have, when x = 1, y = 0. so for y_1 the following happens:

    0 = 1 + a + b

    \Rightarrow a + b = -1 ..............(1)

    similarly, for y_2, the following happens:

    0 = c - 1

    \Rightarrow \boxed {c = 1}


    So we know the value of c, and we know one equation that relates a and b. but what else do we know?

    they share a common tangent at (1,0), in particular, the slope of the tangent line at x = 1 is the same for the two curves

    so, y_1 ' = 2x + a, when x = 1, we get: y_1 ' = 2 + a

    and

    y_2 ' = c - 2x, when x = 1, we get: y_2 ' = c - 2

    Now we equate the two slopes:

    \Rightarrow 2 + a = c - 2

    \Rightarrow a = c - 4 ...............(2)

    \Rightarrow \boxed {a = -3} ............since c = 1

    But from equation (1), a + b = -1

    \Rightarrow b = -1 - a

    \Rightarrow \boxed {b = 2}


    So,  \boxed {y_1 = x^2 -3x + 2} \mbox { and } \boxed {y_2 = x - x^2}
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  3. #3
    MHF Contributor red_dog's Avatar
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    For a and b we can do in this way:
    The two curves have one point of intersection (the tangent point) so the equation x^2+ax+b=cx-x^2 must have one solution (or equal roots). The equation can be written as 2x^2+(a-c)x+b=0.
    Then \triangle =(a-c)^2-8b=(a-1)^2-8b=0.
    Solving the system \left\{\begin{array}{l}a+b=-1\\(a-1)^2-8b=0\end{array}\right. we have a=-3,b=2
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