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Thread: Parabola and Line Intersection Points

  1. #1
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    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
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  2. #2
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    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 $\displaystyle y=\pm 2x - 2$ are tangents to the parabola. So, on that assumption, we can find the tangent at the point $\displaystyle (x_1, kx_1^2)$, and then compare it to one of these equations. Like this:

    $\displaystyle y = kx^2$

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

    So the tangent at this point is

    $\displaystyle y-kx_1^2 =2kx_1(x-x_1)$

    i.e. $\displaystyle y =2kx_1x -kx_1^2$

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

    $\displaystyle 2kx_1 =2$ and $\displaystyle -kx_1^2 = -2$

    $\displaystyle \Rightarrow kx_1= 1$ and $\displaystyle kx_1^2 =2$

    $\displaystyle \Rightarrow k =0.5$ and $\displaystyle x_1 = 2$

    Can you complete it now?

    Grandad
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  3. #3
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    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; Mar 15th 2009 at 02:07 AM.
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  4. #4
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    Parabola

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

    $\displaystyle y = kx^2$

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

    So the tangent at this point is

    $\displaystyle y-kx_1^2 =2kx_1(x-x_1)$

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

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

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

    $\displaystyle \Rightarrow kx_1= 1$ and $\displaystyle kx_1^2 =2$

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

    $\displaystyle \Rightarrow k =0.5$ and $\displaystyle x_1 = 2$

    Can you complete it now?

    Grandad

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

    Is that OK now?

    Grandad
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  5. #5
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    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
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  6. #6
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    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 $\displaystyle y = kx^2$ meets the line $\displaystyle y = 2x - 2$ where

    $\displaystyle kx^2 = 2x - 2$

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

    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 $\displaystyle (-2)^2 - 4\cdot k \cdot 2 = 0$

    $\displaystyle \Rightarrow 8k = 4$

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

    $\displaystyle \Rightarrow x = ...$ etc.

    Grandad
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  7. #7
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    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!
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