Results 1 to 7 of 7

Math Help - Parabola and Line Intersection Points

  1. #1
    Junior Member
    Joined
    Dec 2007
    Posts
    59

    Parabola and Line Intersection Points

    Here is the problem:
    A graphic designer draws a logo involving a parabola ‘sitting’ in a V shape on a set of axes as shown at right.



    Find the equation of the parabola, given it is of the form y = kx2 and the points of intersection of the V with the parabola.

    You probably can't see the picture, so the equations of the lines are -2x-2 and 2x-2, and the only point I have for the parabola is (0,0) as the turning point. I'm having trouble with the first part, finding the equation of the first point, because I can't find k. If I can get this bit, I can work the rest out

    Thanks,
    Steven
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Equation of parabola

    Hello Steven
    Quote Originally Posted by Stevo_Evo_22 View Post
    Here is the problem:
    A graphic designer draws a logo involving a parabola ‘sitting’ in a V shape on a set of axes as shown at right.



    Find the equation of the parabola, given it is of the form y = kx2 and the points of intersection of the V with the parabola.

    You probably can't see the picture, so the equations of the lines are -2x-2 and 2x-2, and the only point I have for the parabola is (0,0) as the turning point. I'm having trouble with the first part, finding the equation of the first point, because I can't find k. If I can get this bit, I can work the rest out

    Thanks,
    Steven
    Although you haven't said so, I'm guessing that the lines y=\pm 2x - 2 are tangents to the parabola. So, on that assumption, we can find the tangent at the point (x_1, kx_1^2), and then compare it to one of these equations. Like this:

    y = kx^2

    \Rightarrow \frac{dy}{dx}= 2kx =2kx_1 at (x_1, kx_1^2)

    So the tangent at this point is

    y-kx_1^2 =2kx_1(x-x_1)

    i.e. y =2kx_1x -kx_1^2

    So if this is the same as the line y = 2x -2, we can compare coefficients to get:

    2kx_1 =2 and -kx_1^2 = -2

    \Rightarrow kx_1= 1 and kx_1^2 =2

    \Rightarrow k =0.5 and x_1 = 2

    Can you complete it now?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2007
    Posts
    59
    Ok now I'm really lost Sorry I don't know when you're talking about the parabola or the line...
    Last edited by Stevo_Evo_22; March 15th 2009 at 03:07 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Parabola

    Hello Steven
    Quote Originally Posted by Grandad View Post
    Although you haven't said so, I'm guessing that the lines y=\pm 2x - 2 are tangents to the parabola. So, on that assumption, we can find the tangent at the point (x_1, kx_1^2), and then compare it to one of these equations. Like this:

    y = kx^2

    \Rightarrow \frac{dy}{dx}= 2kx =2kx_1 at (x_1, kx_1^2)

    So the tangent at this point is

    y-kx_1^2 =2kx_1(x-x_1)

    i.e. y =2kx_1x -kx_1^2
    This is the equation of the tangent to the parabola at any general point \color{red}(x_1, kx_1^2) on the parabola. Do you understand how I got this?

    So if this is the same as the line y = 2x -2 (i.e. one of the two lines you were given in the question), we can compare coefficients to get:

    Coefficients of \color{red}x in the two equations are the same: 2kx_1 =2 and the constant terms are the same: -kx_1^2 = -2

    \Rightarrow kx_1= 1 and kx_1^2 =2

    Divide the second equation by the first: \color{red}\frac{kx_1^2}{kx_1} = \frac{2}{1}

    \Rightarrow k =0.5 and x_1 = 2

    Can you complete it now?

    Grandad

    I've put further explanations in
    red. What this shows is that the parabola y = 0.5x^2 has a tangent at the point (2, 2) whose equation is y = 2x -2; i.e. one of the lines that you were given. Because the parabola is symmetrical about the y-axis, the tangent at (-2,2) will be the other line you were given: y = -2x -2.

    Is that OK now?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2007
    Posts
    59
    Thankyou very much for your help, and sorry I couldn't get back to you sooner

    Unfortunately I still do not understand how you did it, because I've never seen something like that before, however I was able to work out the question using the discriminant.

    Thanks again,
    Steven
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Parabola

    Hello Steven

    Sorry if my explanation used stuff you haven't covered in your course. I assumed that you had done some calculus, which it now appears you haven't.

    Yes, you can solve the problem without calculus using the discriminant of a quadratic equation. I imagine you said something like this:

    The curve y = kx^2 meets the line y = 2x - 2 where

    kx^2 = 2x - 2

    \Rightarrow kx^2 - 2x +2 = 0

    The line is a tangent to the curve if the roots of this equation are equal. (This is the hard bit - do you understand it?)

    i.e. if (-2)^2 - 4\cdot k \cdot 2 = 0

    \Rightarrow 8k = 4

    \Rightarrow k = 0.5, which is the same answer as I got using calculus methods.

    \Rightarrow x = ... etc.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2007
    Posts
    59
    Yes you're correct-I have yet to do calculus

    So as you said, the line only touches the curve so the discriminant must equal 0 for there to only be one solution. All I had to do was simply solve for k as you did with calculus...

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 19th 2011, 09:11 AM
  2. Points of intersection, circle and parabola
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: July 10th 2010, 05:12 AM
  3. Replies: 1
    Last Post: October 20th 2009, 02:52 AM
  4. Replies: 2
    Last Post: March 5th 2009, 09:45 AM
  5. Replies: 1
    Last Post: February 12th 2007, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum