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Math Help - Help: triangle under parabola

  1. #1
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    Help: triangle under parabola

    I'm not sure if this is algebra or geometry; I haven't studied math in a very long time. Anyway, I really need help with this problem. I've attached an image of the problem, and here's a translation of the text from Spanish:

    An isosceles rectangular triangle is inscribed in the parabola y^2=4x with the right angle in the vertex of the curve like shows the figure. The area of the triangle AVC is: (as shown)

    I've already spent about an hour tearing my hair out and looking through books to try to figure out how to do this. Any help would be greatly appreciated. Thanks a bunch
    Attached Thumbnails Attached Thumbnails Help: triangle under parabola-problem.gif  
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  2. #2
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    Quote Originally Posted by mpoderoso View Post
    I'm not sure if this is algebra or geometry; I haven't studied math in a very long time. Anyway, I really need help with this problem. I've attached an image of the problem, and here's a translation of the text from Spanish:

    An isosceles rectangular triangle is inscribed in the parabola y^2=4x with the right angle in the vertex of the curve like shows the figure. The area of the triangle AVC is: (as shown)

    I've already spent about an hour tearing my hair out and looking through books to try to figure out how to do this. Any help would be greatly appreciated. Thanks a bunch
    The base of the triangle is 2q and the height is p.

    The area of a triangle is \frac{1}{2}\times\textrm{ base }\times\textrm{ height}

     = \frac{1}{2}\times 2q\times p

     = pq.
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  3. #3
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    Is this supposed to be a multiple choice? The image gives me that impression, but none of the answers make sense.

    You can substitute p and q into the parabola's equation and end up with q = \pm \sqrt{4p}, and you can solve for p as p = \frac {q^2}{4}. As Prove It says, the area is clearly pq. We can substitute and get the area in terms of one or the other variable and end up with: A = 2p^{\frac {3}{2}} and A = \frac {q^3}{4}.

    Anyways, I tried to see which of those formulas in the image gave the correct answer, and none of them seem to. For example, with p = 9, we have q = 6, giving an area of 54. None of those formulas gives 54 as the answer.

    So if that is multiple choice, either I'm confused or they are all incorrect.
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  4. #4
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    Hello, mpoderoso!

    Are you sure of the translation?
    None of the answers makes sense.
    In fact, none of it makes any sense.


    An isosceles right triangle is inscribed in the parabola y^2\,=\,4x
    with the right angle in the vertex of the curve like shows the figure.

    The area of the triangle AVC is:

    . . (A)\;\frac{q^2+4q}{2} \qquad(B)\;p^2+2p \qquad(C)\;\frac{p^2+4p}{2} \qquad (D) \;q^2+2q .??


    Code:
            |         A
            |         * (p,q)
            |   *   / |
            | *   /   | q
            |*  /     |
            | /       |
          V * - - - - + - - - -
            | \   p   |
            |*  \     |
            | *   \   | q
            |   *   \ |
            |         * (p,-q)
            |         C
            |

    The base of the triangle is: AC \,=\,2q
    The height of the triangle is: p

    The area is: . A \;=\;\tfrac{1}{2}\text{(base)(height)} \;=\;\tfrac{1}{2}(2q)(p) \quad\Rightarrow\quad\boxed{A\;=\;pq}



    Since they gave us the equation of the parabola: . y^2 \,=\,4x
    . . we can locate vertices A and C exactly.

    Sides VA and VC are "45-lines".
    . . Their equations are: . y = x and y = -x
    They intersect the parabola at: .(4,4) and (4,-4).

    . . Therefore, the base is 8 and the height is 4 . . . The area is {\color{blue}16}.

    So what's with the p's and q's ?

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