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Math Help - equation tanget of parabola

  1. #1
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    equation tanget of parabola

    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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    It is possible that there is an easier way to do this, but anyway:

    Let the common tangent meet the first curve at (x_1,y_1) and the second curve at (x_2,y_2). Then the derivatives of those curves at those points are equal. Call this gradient m. then y_1+m(x_2-x_1) = y_2 (draw a diagram if you are unsure of this one). This gives
    5 equations for 5 unknowns, which surprisingly solve quite easily. Remember that you only need one point and m to find the equation of the tangent.
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    Please tell me if what i have done is correct

    Sir i have tried to solve this question as per ur directions .pls tell me if it is correct


    Equation of 1st parabola is  y^2 = 4ax
    Let the point of tangent of the parabola be (x_1,y_1)
    Therefore
    Eq of tangent to the parabola will be yy_1 = 2a(x+x_1)

    Similarly for the parabola x^2=4by
    Eq of tangent to the 2nd parabola will be xx_2 = 2b(y+y_2)

    Please tell me what needs to be done next ..........
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  4. #4
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    Quote Originally Posted by Shivanand View Post
    Find the equation of the common tangent to the parabolas y^2 = 4ax and x^2=4by

    The common tangent is found by the derivatives being equal.

    So we'd need to write both equations as y in terms of x.

    The equations are y = 2\sqrt{a}\sqrt{x} and y = \frac{1}{4b}x^2.

    Find their derivatives and set them equal to each other...

    \frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x.

    Solve for x.

    \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x

    2b\sqrt{a}= x^{\frac{3}{2}}

    x = 2^{\frac{2}{3}}b^{\frac{2}{3}}a^{\frac{1}{3}}.


    Substitute this into one of the equations to find y.

    Can you go from here?
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  5. #5
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    Common tangents to parabolas

    Hi -

    Quote Originally Posted by Prove It View Post
    Find their derivatives and set them equal to each other...

    \frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{x}} = \frac{1}{2b}x.

    Solve for x.
    Sorry, I'm afraid all this does is to find the value of x - the same value on each curve - at which the tangents are parallel.

    Can I suggest:

    Re-write x^2=4by as y=\frac{x^2}{4b}

    \therefore \frac{dy}{dx} = \frac{x}{2b}

    When x=k, y=\frac{k^2}{4b} and \frac{dy}{dx} = \frac{k}{2b}

    So the tangent to this curve at the point where x = k is:

    y - \frac{k^2}{4b}= \frac{k}{2b}(x-k)

    i.e. x=\frac{4by+k^2}{2k}

    This meets y^2=4ax where

    y^2=2a \left(\frac{4by+k^2}{2k}\right)

    Re-arrange as a quadratic in y:

    ky^2-8aby-2ak^2=0

    Now this line is also a tangent to y^2=4ax if this quadratic has equal roots.

    So ...

    (I'll leave you to do the next few steps.)

    The equation turns out to be:

    b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3  }}=0

    Grandad
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    Please tell if this is right

    I have tried it the way u told me pls tell me if this is right

    y^2 = 4ax \Rightarrow[ math]y=2\sqrt{a}\sqrt{x}[/tex] \rightarrow(1)

    x^2=4by \Rightarrow y=\frac{x^2}{4b} \rightarrow(2)

    Differetiating eq 1 wrt to x weget
    \frac{dy}{dx}= \frac{d}{dx} (2\sqrt{a}\sqrt{x})

    = 2\sqrt{a}\frac{d}{dx} (x)^\frac{1}{2}

    = 2\sqrt{a}.\frac{1}{2\sqrt{x}} .\frac{d}{dx} (x)

    = \frac{\sqrt{a}}{\sqrt{x}} (1)

    = \frac{\sqrt{a}}{\sqrt{x}} \rightarrow(3)

    Similarly for eq 2

    \frac{dy}{dx}= \frac{d}{dx} [\frac{x^2}{4b}]

    = \frac{1}{4b} \frac{d}{dx} x^2

    = \frac{1}{4b} 2x

    = \frac{x}{2b} \rightarrow(4)

    Setting them equal to each other weget

    \frac{\sqrt{a}}{\sqrt{x}}= \frac{x}{2b}

    2b\sqrt{a}= x\sqrt{x}

    2b\sqrt{a}= x.x^\frac{1}{2}

    2b\sqrt{a}= x^\frac{3}{2}

    2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{2} ^\frac{2}{3}= x

    2^\frac{2}{3} b^\frac{2}{3} a^\frac{1}{3}=x

    Therefore x = 2^\frac{2}{3} b^\frac{2}{3} a^\frac{1}{3}

    Now substituting x in one of the equation y^2=4ax weget

    y^2=4ax

    y^2 = 4a(2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})

    y= \sqrt{4}\sqrt{2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{  4}{3}}

    y= 2(2^\frac{2}{3} .b^\frac{2}{3}.a^\frac{4}{3})^\frac{1}{2}

    y= 2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})

    Now the equation of tangent to parabola y^2=4ax at point (x_1,y_1) would be
    yy_1=2a(x+x_1)

    ie
    y.2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})= 2a(x+2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})
    Last edited by varunnayudu; December 17th 2008 at 02:13 AM.
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  7. #7
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    Is it right?

    Hi-

    Quote Originally Posted by varunnayudu View Post
    pls tell me if this is right
    No, sorry, it's not right at all. Please see my previous posting.

    Grandad
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  8. #8
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    Hw did u get this answer.

    i.e.

    This meets where



    sir hw did u get this answer

    when y^2 = 4ax

    and x = \frac{4by+k^2}{2k}

     <br />
y^2 = 4a (\frac{4by+k^2}{2k})<br />


    Hw did u get

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  9. #9
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    Typo!

    Hello varunnayudu
    Quote Originally Posted by varunnayudu View Post
    sir hw did u get this answer

    when y^2 = 4ax

    and x = \frac{4by+k^2}{2k}

     <br />
y^2 = 4a (\frac{4by+k^2}{2k})<br />


    Hw did u get

    Oops! Typo. I copied and pasted, and forgot to remove the 2 from the denominator. Of course, it should read:

     <br />
y^2 = 2a (\frac{4by+k^2}{k})<br />

    The next line of my posting is correct, though.
    Re-arrange as a quadratic in :



    Let me complete the working for you.

    Now this line is also a tangent to if this quadratic has equal roots; i.e. if

    (-8ab)^2 = 4k(-2ak^2)

    \implies -8ak^3 = 8^2a^2b^2

    \implies k^3=-8ab^2

    \implies k=-2a^{\frac{1}{3}}b^{\frac{2}{3}}

    Re-arrange the equation of the tangent found previously to read:

    4by -2kx + k^2=0

    and then substitute this value of k into this equationto get:

    4by +4a^{\frac{1}{3}}b^{\frac{2}{3}}x + 4a^{\frac{2}{3}}b^{\frac{4}{3}}=0

    i.e. b^{\frac{1}{3}}y+a^{\frac{1}{3}}x + (ab)^{\frac{2}{3}}=0

    I hope that you understand this now.

    Grandad
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  10. #10
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    Hw did u get this

    Sir,


    The next line of my posting is correct, though. Re-arrange as a quadratic in :



    Let me complete the working for you.

    Now this line is also a tangent to
    if this quadratic has equal roots; i.e. if



    ---------------

    Hw did u substitute y =-8ab and x = -2ak^2

    specify this
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  11. #11
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    And Hw did u get this



    Re-arrange the equation of the tangent found previously to read:




    ----------------------------------------

    which eq do i have to substitute to get that eq as stated above
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  12. #12
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    Equal roots of a quadratic

    Hello varunnayudu
    Quote Originally Posted by varunnayudu View Post
    The next line of my posting is correct, though. Re-arrange as a quadratic in :



    Let me complete the working for you.

    Now this line is also a tangent to
    if this quadratic has equal roots; i.e. if



    ---------------

    Hw did u substitute y =-8ab and x = -2ak^2

    specify this

    The quadratic
    ax^2 +bx+c=0 has equal roots if b^2=4ac.

    So

    has equal roots if:

    (-8ab)^2 = 4k(-2ak^2)

    Grandad
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  13. #13
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    Equation of common tangent

    Hello varunnayudu
    Quote Originally Posted by varunnayudu View Post

    Quote Originally Posted by varunnayudu View Post
    Re-arrange the equation of the tangent found previously to read:




    ----------------------------------------

    which eq do i have to substitute to get that eq as stated above

    Here's a summary of how we solved this question.

    By differentiation, we found that the gradient of x^2=4by
    when x=k is \frac{k}{2b}, and the value of y at this point is \frac{k^2}{4b}.

    Then, using y - y_1 = m(x-x_1)
    we found the tangent to x^2=4by at the point (k, \frac{k^2}{4b}) is

    y-\frac{k^2}{4b}=\frac{k}{2b}(x-k)

    Re-arranging and simplifying this equation we get:

    4by -2kx +k^2=0 (1)

    Next, we found the value of k
    that made this line a tangent to y^2=4ax, using the fact that a tangent meets a curve at two coincident points, and therefore the quadratic equation giving the values of y has equal roots.

    Then into equation (1) we substitute this value of k (= -2a^{\frac{1}{3}}b^{\frac{2}{3}})
    to obtain the tangent that is common to both parabolas. When we do this, and simplify the result, we get

    b^{\frac{1}{3}}y+a^{\frac{1}{3}}x+(ab)^{\frac{2}{3  }}=0

    No more questions about this problem now, please.

    Grandad

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