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Math Help - Tangent to a parabola through a given point.

  1. #1
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    Tangent to a parabola through a given point.

    Hi guys
    can you please help me with the following question

    I really don't how to start with this question
    Thanks
    Last edited by mr fantastic; August 14th 2010 at 06:57 PM. Reason: Re-titled.
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  2. #2
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    Well, set up your favorite form of an equation of a line. It should probably have an x, a y, and two or three more variables (parameters) in it. The fact that the line passes through (2,-3) should give you an equation and/or some other fact about the parameters. The fact that the line is tangent to that parabola should give you more information.

    Alternatively, find the tangent line to the parabola at a general point, and see which of those lines pass through (2,-3).
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  3. #3
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    I tried to solve the problem and I got the equation for the first line y=-x-1 which i guess is right, but i could not find the equation of the second line... can you help me
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    Quote Originally Posted by metallica007 View Post
    I tried to solve the problem and I got the equation for the first line y=-x-1 which i guess is right, but i could not find the equation of the second line... can you help me
    Equation of tangent at any point on the given parabolla is given by

    \frac{y + y_1}{2} = xx_1 + \frac{x + x_1}{2}

    This tangent passes through (2, -3). Put x = 2 and y = -3 in the above equation and solve for y1. Put this value in the equation of the aprabolla and solve for x1.

    You get two values for x1. Find the corresponding values for y1.

    Substitute these values in the equation of tangent.
    Last edited by sa-ri-ga-ma; August 14th 2010 at 11:05 PM.
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  5. #5
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Equation of tangent at any point on the given parabolla is given by

    \frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2}

    This tangent passes through (2, -3). Put x = 2 and y = -3 in the above equation and solve for y1. Put this value in the equation of the aprabolla and solve for x1.

    You get two values for x1. Find the corresponding values for y1.

    Substitute these values in the equation of tangent.
    this is the first time i knew about the equetion \frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2} !!
    guys i am completely lost, can you please solve the problem for me please ..
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  6. #6
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    Quote Originally Posted by metallica007 View Post
    this is the first time i knew about the equetion \frac{y - y_1}{2} = xx_1 + \frac{x + x_1}{2} !!
    guys i am completely lost, can you please solve the problem for me please ..
    Sorry. There is typo.

    It should be (y+y1)/2.
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  7. #7
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    The given parabola is y= x^2+ x which has derivative 2x_0 when x= x_0. The slope of any tangent to that parabola must be 2x_0 for some number x_0.

    Also, any line through (2, 3) is of the form y- 3= m(x- 2) so a line through (2, 3) tangent to the parabola is of the form y- 3= (2x_0)(x- 2).

    Of course, the tangent line must touch the parabola at x= x_0, that is when x= x_0, y= x_0^2+ x_0

    That is, you must have x_0^2+ x_0- 3= 2x_0(x_0- 2). Solve that for x_0 and then write the equation of the tangent lines. (That is a parabola so there will be two solutions.)
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  8. #8
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    Quote Originally Posted by metallica007 View Post
    Hi guys
    can you please help me with the following question

    I really don't how to start with this question
    Thanks
    Here comes a slightly different approach to this problem:

    1. A line passing through (2, -3) has the equation:

    y-(-3)=m(x-2)~\implies~y=mx-2m-3

    3. Calculate the coordinates of the points of intersection of the parabola and the line:

    x^2+x=mx-2m-3~\implies~x^2+(1-m)x+2m+3=0

    Solve for x:

    x = \dfrac{-(1-m) \pm \sqrt{(1-m)^2-4 \cdot 1 \cdot (2m+3)}}{2}

    x = \dfrac{m-1}2 \pm \dfrac12 \cdot \sqrt{m^2-10m-11}

    3. A line is tangent to the parabola if there exist only one point of intersection: the tangent point. This is only possible if the dicriminant equals zero:

    m^2-10m-11=0~\implies~m=-1~\vee~m=11

    Plug in these values into the equation of the line at 2.

    4. The x-coordinate of the tangent point is \dfrac{m-1}2. Plug in the values of m to get the x-coordinates and plug in the x-coordinates into the equation of the line to get the corresponding y-coordinates of the tangent points.

    5. For confirmation draw a sketch of the parabola and the two lines. (My sketch isn't drawn to scale!)
    Attached Thumbnails Attached Thumbnails Tangent to a parabola through a given point.-tang_parab.png  
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