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Math Help - Area of triangle in 2D space

  1. #1
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    Area of triangle in 2D space

    Hi,

    I have a problem I'm having trouble with.

    The points P(ap^2, 2ap) and Q(aq^2, 2aq) lie on the parabola with equation y^2 = 4ax. Show that the tangents to the parabola at P and Q meet at R[apq, a(p+q)]. Show further that the area of the triangle PQR is \frac{1}{2}a^2 |p-q|^3

    I can do the first part about showing R, but I have no idea about finding the area of the triangle. Any help would be much appreciated

    Stonehambey
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  2. #2
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    Hello, Stonehambey!

    The points P(ap^2, 2ap) and Q(aq^2, 2aq) lie on the parabola with equation y^2 = 4ax.

    (a) Show that the tangents to the parabola at P and Q meet at R[apq, a(p+q)].

    (b) Show that the area of \Delta PQR is: . \tfrac{1}{2}a^2 |p-q|^3
    Lucky for us, there is a formula for the area . . .


    The area of the triangle with vertices: (x_1,y_1),\:(x_2,y_2),\:(x_3,y_3)

    . . . . . is given by: . A \;=\;\frac{1}{2}\left|\begin{array}{ccc}x_2 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3& y_3 & 1 \end{array}\right|

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  3. #3
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    A specific case that can be generalized: q<0<p

    Forgive me, I don't know any fancy geometry formulas for area, but this one has a nice graphical interpretation...

    Without loss of generality, let p>0 and q<0 (we know a>0). The graph is a parabola opening from the origin to the right. P sits at an arbitrary point at the top, Q on the bottom. R is somewhere off in the 2nd or 3rd quadrant near the x-axis...

    Cut the triangle in half and figure the areas of each separately. Draw a horizontal line through R at y=a(p+q). Where does it intersect PQ?

    Use the point-slope form of a line to connect P and Q via y=2/(p+q)*(x-ap^2)+2ap. Substitute a(p+q) in for y and solve for x... x=a(p^2+q^2)/2, call this point M (NOT necessarily the midpoint of P and Q).

    So the "base" of both triangles is segment RM, length a(p^2+q^2)/2 - apq, simplifies to |RM|=a(p-q)^2/2.

    The height of triangle RPM is |2ap-2a(p+q)|=-2aq (remember q is negative by definition)
    The height of triangle RQM is |2aq-2a(p+q)|=2ap

    So the area of RPM= -q*a^2(p-q)^2/2
    And the area of RQM= p*a^2(p-q)^2/2

    Whose sum is (1/2)a^2(p-q)^3

    NOTE: THIS STILL WORKS IF WE ACCEPT THE IDEA OF "NEGATIVE AREA." IF p and q ARE BOTH POSITIVE OR BOTH NEGATIVE, THESE NUMBERS STILL WORK. EX: IF p=q, ACCORDING TO THE ABOVE EQN, AREA=0, AS EXPECTED. WE WOULD SIMPLY FIND THE AREA OF THE BIGGER HALF-TRIANGLE, AND SUBTRACT OFF THE AREA OF THE SMALLER HALF-TRIANGLE, SINCE THEY WOULD BE OVERLAPPING INSTEAD OF APPENDING EACH OTHER.
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  4. #4
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    Formula works nicely. For the geometric interpretation, I was trying (without luck) a similar approach. So I'll keep reading over it until it's completely clear to me

    thanks guys
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