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Isocleles Triangles in a Parabola Help

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Triangle OAB is an isosceles triangle with vertex O at the origin and vertices A and B on the parabola y = 9-x^2
    Express the area of the triangle as a function of the x-coordinate of A.


    2. Relevant equations

    A = 1/2 bh
    Distance formula (maybe)
    Heron's Formula (an alternative)


    3. The attempt at a solution

    The area of a triangle is given by A = 1/2 bh. If point A is (x, 9 - x^2), then b=2x and h = 9-p^2
    The final equation will be A = 9x - x^3 (as a function of the x coordinate of A)

    The equation above works only for a limited number of triangles embedded in the parabola. For instance, if A and B has the same x coordinates, then the equation will work. But i considered two different triangles that can make an isosceles triangle:

    O = (0,0)
    A = (2,5)
    B = (2,5)
    Area is 10 unites^2

    and

    O = (0,0)
    A = (2,5)
    B = (-2.2777902,3.8111609)
    And the area will be around 4.75 units^2

    Both of those are isosceles triangles, that can fit into the parabola of 9-x^2. How would I go about finding an equation that can find the area of all possible isosceles triangles in the parabola?
     
    Last edited: Sep 3, 2007
  2. jcsd
  3. Sep 3, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You mean 9- x2, of course.

    Yes, if OA and OB are the two congruent sides then the problem is easy as you say. Suppose, instead, that OA and AB are the conguent sides. Let (x0,y0) be the point A (and we may, without loss of generality, assume that x0 is negative. B= (x,y) must satisfy y= 9- x2 and [itex](x-x_0)^2+ (y-y_0)^2= x^2+ y^2[/itex] If you multiply out the left side, you will see that the x2 and y2 terms cancel leaving [itex]-2x_0x+ x_0^2- 2y_0y+ y_0^2= 0[/itex] or [itex](2x_0)x+ (2y_0)y= x_0^2+ y_0^2[/itex]. It should not be too difficult to solve those two equations for x and y and then calculate the area of the triangle in terms of x0 and y0.

    The case where OB and AB are the congruent sides is just the reflection of the previous case in the y-axis and gives the same area formula.
     
  4. Sep 4, 2007 #3
    Another way is, that since the triangle is isosceles, the distance of the two points will be the same from y axis. If the coordinates of A are (x,y), the coordinates of B will be (-x,y). So, one coordinate will be (0,9) [vertex], (x,y) [A], and (-x,y) .

    Now, we know that [tex]y=9-x^2[/tex] and the area of a triangle can be given by the determinant, [tex]D=\frac{1}{2}(0 9 1, x 9-x^2 1, -x 9-x^2, 1)[/tex]. (Sorry I dont know the tex for a determinant, hopefully someone who does could format it. Thank you.)

    Solving this should be easy, specially if you use the properties of a det.
     
  5. Sep 4, 2007 #4

    HallsofIvy

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    Staff Emeritus
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    But the real question was about the case where the "base" of the triangle, the non-congruent line- was NOT parallel to the x-axis.
     
  6. Sep 4, 2007 #5
    Sorry. My mistake. My case only works for the base parallel to the x axis.
     
  7. Sep 4, 2007 #6
    For the other case, if you assume the coordinates of B to be [tex](x_2,y_2)[/tex] and of A to be [tex](x_1,y_1)[/tex] and since the vertex is at the origin, assuming OB to be the base, OA comes out to be [tex]\sqrt{x_1^2+y_1^2}[/tex].

    The equation of the line OA comes out to be [tex]y=(\frac{y_1}{x_1})x[/tex]. The distance of the point B from OA is [tex]\frac{y_2-\frac{y_1}{x_1}x_2}{\sqrt{1+\frac{y_1^2}{x_1^2}}}[/tex].

    After solving for [tex]\frac{1}{2}\frac{y_2-\frac{y_1}{x_1}x_2}{\sqrt{1+\frac{y_1^2}{x_1^2}}}\sqrt{x_1^2+y_1^2}[/tex]

    The area is given by [tex]\frac{1}{2}(x_1y2-y_1x_2)[/tex]. If you could find a relation between [tex]x_1, y_1[/tex] and [tex]x_2,y_2[/tex] you would be home free... how would you do that though?
     
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