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Math Help - Find an equation of tangent

  1. #1
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    Find an equation of tangent

    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^=4ay.


    Please help me to solve this.
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  2. #2
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    I've done this
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  3. #3
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    Quote Originally Posted by geton View Post
    I've done this
    Why don't you share? Your solution might help someone else.

    -Dan
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    Quote Originally Posted by geton View Post
    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^=4ay.

    Please help me to solve this.
    Quote Originally Posted by topsquark View Post
    Why don't you share? Your solution might help someone else.

    -Dan
    Let me try

    y^2 = 4ax \Rightarrow(x^2/4a)^2 = 4ax \Rightarrow x^4 = 64xa^3 \Rightarrow x = 0, \pm 4a
    Clearly -4a wont do. So the intersection points are (0,0),(4a,4a)

    The tangent at (4a,4a)

    \frac{dy}{dx}\bigg{|}_ {x=4a} = \frac{d(\frac{x^2}{4a})}{dx}\bigg{|}_ {x=4a} = \frac{x}{2a}\bigg{|}_ {x=4a} = 2

    y - 4a = \frac{dy}{dx}\bigg{|}_ {x=4a}(x - 4a) \Rightarrow y - 4a = 2(x - 4a) \Rightarrow y = 2x - 4a
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Why don't you share? Your solution might help someone else.

    Let the tangent is y = mx + c

    Substitute this into parabola's equation with y^2 = 4ax.
    So: (mx + c)^2 = 4ax

    There is only one root because tangent meets parabola at only one point.

    From this equation we found mc = a


    Same thing for x^2 = 4ay.


    By solving two equations simultaneously, we get the tangent equation,
    y + x +a = 0.
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