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Math Help - area between the parabola

  1. #1
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    area between the parabola

    find the area between the parabola y^2 = 4ax and the line y= mx......... even find the equation of the common tangent to the parabolas y^2=4ax and x^2=4by........
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  2. #2
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    Area between y^2 = 4ax and y = mx

    Substitute x = \frac{y}{m} so we can find the intersection points which we will use for the limits of integration:

    y^2 = 4a \times \frac{y}{m}

    y^2 = \frac{4ay}{m}

    y(y - \frac{4a}{m}) = 0

    y = 0 or y = \frac{4a}{m}

    Area = \int_0^{\frac{4a}{m}}\, \left( \frac{y}{m} - \frac{y^2}{4a}\right)\, dy

    Can you solve this now?
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  3. #3
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    Quote Originally Posted by nzmathman View Post
    Area between y^2 = 4ax and y = mx

    Substitute x = \frac{y}{m} so we can find the intersection points which we will use for the limits of integration:

    y^2 = 4a \times \frac{y}{m}

    y^2 = \frac{4ay}{m}

    y(y - \frac{4a}{m}) = 0

    y = 0 or y = \frac{4a}{m}

    Area = \int_0^{\frac{4a}{m}}\, \left( \frac{y}{m} - \frac{y^2}{4a}\right)\, dy

    Can you solve this now?
    Alternatively, in order to get it in familiar terms (i.e. in terms of x), substitute  y = mx and find the intersection points.

    I.e. (mx)^2 = 4ax

    m^2x^2 - 4ax = 0

    x(m^2 x - 4a) = 0

     x= 0 or x = \frac{4a}{m^2}.


    So Area = \int_0^{\frac{4a}{m^2}}{(mx - 2\sqrt{a}\sqrt{x})\,dx}.
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  4. #4
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    Yes that's what I originally started out doing, but was trying to avoid integration of the square root.
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    Quote Originally Posted by nzmathman View Post
    Yes that's what I originally started out doing, but was trying to avoid integration of the square root.
    Why?

    Integrating a square root is easy...

    Remember that \sqrt{x} = x^{\frac{1}{2}}


    So \int{\sqrt{x}\,dx} = \int{x^{\frac{1}{2}}\,dx}

     = \frac{1}{\frac{1}{2}}x^{\frac{1}{2}-1} + C

     = 2x^{-\frac{1}{2}} + C

     = \frac{2}{x^{\frac{1}{2}}} + C

     = \frac{2}{\sqrt{x}} + C

     = \frac{2\sqrt{x}}{x} + C.


    You just need to remember your index laws
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  6. #6
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    Actually,

    \int \sqrt{x}  dx = \frac{2}{3} x^{\frac{3}{2}} + C= \frac{2}{3} x \sqrt{x} + C



    I know it's easy but I was trying to make the integration a little easier for the person who posed the question.
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    Quote Originally Posted by nzmathman View Post
    Actually,

    \int \sqrt{x}  dx = \frac{2}{3} x^{\frac{3}{2}} + C= \frac{2}{3} x \sqrt{x} + C



    I know it's easy but I was trying to make the integration a little easier for the person who posed the question.
    Oh - oops! My bad lol... Yes you are right :P
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    Common tangents to parabolas

    Hi-

    Quote Originally Posted by Shivanand View Post
    ... even find the equation of the common tangent to the parabolas y^2=4ax and x^2=4by........
    Have you not looked at my solution to your earlier posting? http://www.mathhelpforum.com/math-he...-parabola.html

    I promise you - it is correct!

    Grandad
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