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Thread: 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 $\displaystyle (x_1,y_1)$ and the second curve at $\displaystyle (x_2,y_2)$. Then the derivatives of those curves at those points are equal. Call this gradient m. then $\displaystyle y_1+m(x_2-x_1) = y_2$ (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 $\displaystyle y^2 = 4ax $
    Let the point of tangent of the parabola be $\displaystyle (x_1,y_1)$
    Therefore
    Eq of tangent to the parabola will be $\displaystyle yy_1 = 2a(x+x_1)$

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

    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 $\displaystyle y = 2\sqrt{a}\sqrt{x}$ and $\displaystyle y = \frac{1}{4b}x^2$.

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

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

    Solve for x.

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

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

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


    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...

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

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

    Can I suggest:

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

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

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

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

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

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

    This meets $\displaystyle y^2=4ax$ where

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

    Re-arrange as a quadratic in $\displaystyle y$:

    $\displaystyle ky^2-8aby-2ak^2=0$

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

    So ...

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

    The equation turns out to be:

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

    Grandad
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  6. #6
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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

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

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

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

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

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

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

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

    Similarly for eq 2

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

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

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

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

    Setting them equal to each other weget

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

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

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

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

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

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

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

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

    $\displaystyle y^2=4ax$

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

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

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

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

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

    ie
    $\displaystyle y.2(2^\frac{1}{3}.b^\frac{1}{3}.a^\frac{2}{3})$=$\displaystyle 2a(x+2^\frac{2}{3}.b^\frac{2}{3}.a^\frac{1}{3})$
    Last edited by varunnayudu; Dec 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 $\displaystyle y^2 = 4ax$

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

    $\displaystyle
    y^2 = 4a (\frac{4by+k^2}{2k})
    $


    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 $\displaystyle y^2 = 4ax$

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

    $\displaystyle
    y^2 = 4a (\frac{4by+k^2}{2k})
    $


    Hw did u get

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

    $\displaystyle
    y^2 = 2a (\frac{4by+k^2}{k})
    $

    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

    $\displaystyle (-8ab)^2 = 4k(-2ak^2)$

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

    $\displaystyle \implies k^3=-8ab^2$

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

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

    $\displaystyle 4by -2kx + k^2=0$

    and then substitute this value of $\displaystyle k$ into this equationto get:

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

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

    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 $\displaystyle y =-8ab$ and $\displaystyle 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 $\displaystyle y =-8ab$ and $\displaystyle x = -2ak^2 $

    specify this

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

    So

    has equal roots if:

    $\displaystyle (-8ab)^2 = 4k(-2ak^2)$

    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 $\displaystyle x^2=4by$
    when $\displaystyle x=k$ is $\displaystyle \frac{k}{2b}$, and the value of $\displaystyle y$ at this point is $\displaystyle \frac{k^2}{4b}$.

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

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

    Re-arranging and simplifying this equation we get:

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

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

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

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

    No more questions about this problem now, please.

    Grandad

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