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Tangent-lines of Parabolas

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Tangent lines T1 and T2 are drawn at two points P1 and P2 on the parabola y=x^2 and they intersect at a point P. Another tangent line T is drawn at a point between P1 and P2; it intersects T1 at Q1 and T2 at Q2. Show that:

    (PQ1/PP1) + (PQ2/PP2) = 1


    2. Relevant equations



    3. The attempt at a solution
    Let f(x) function y=f(x)=x^2. Therefore:
    - P1(p1,f(p1))
    - P2(p2,f(p2))
    and the x-coordinate of P is:
    - P: x=(p1+p2)/2

    I've constructed a third point M where the 3rd tangent line T is drawn, therefore:
    - M(m,f(m))

    The x-coordinates of Q1 and Q2 are:
    - Q1: x=(m+p1)/2
    - Q2: x=(m+p2)/2

    I realize that the tangent-lines of any two points on a parabola will intersect at a point whose x-coordinate is the midpoint of the x-coordinates of the two points on the parabola, therefore the x-coordinates of point P will be halfway between points P1 and P2. And the tangent-line T, which corresponds to point M, if placed directly between points P1 and P2, then; (PQ1/PP1) = 1/2 and (PQ2/PP2) = 1/2, therefore the sum is obviously 1 and the x-coordinates of the point Q1 will be directly between P1 and P and the x-coordinate of Q2 will be directly between P2 and P. What i'm having problems doing is proving the statement it asks me to prove and how calculus is involved in this problem. I've thought about using the distance formula but it expands into a very messy problem. I don't understand how to express the line-segments PQ1, PP1, PQ2, PP2 and how it relates to differentiation.
     
    Last edited: Jul 11, 2009
  2. jcsd
  3. Jul 11, 2009 #2
    Hi Samuel,

    I'll give you some hints on a strategy you could use. I've sketched it out on paper and it seems to work out nicely enough.

    Step 1) Choose your 3 points, P1, P2, M and write out their coordinates (i.e. [tex]P_1 = (p_1, p_1^2)[/tex]).

    Step 2) Write out the formula for the tangent lines to [tex]f(x)=x^2[/tex] that pass through your points P1, P2, and M.

    Hint: You want to calculate 3 lines, [tex]y_1, y_2, y_3[/tex]. You will need the slope of each line, and the y-intercept. For example you should get something like [tex]y_1 = (2p_1)x - p_1^2[/tex] for the tangent line that passes through P1.

    Step 3) Calculate the intersection points of [tex](y_1, y_2)[/tex], [tex](y_2, y_3)[/tex], and [tex](y_1, y_3)[/tex].

    Hint: For example, to get the intersection points of [tex]y_1, y_2[/tex], just set [tex]y_1 = y_2[/tex] to find the x coordinate (as you've done above), then plug this x coordinate back into either of [tex]y_1, y_2[/tex] to get the y coordinate.

    Notice you now have explicit coordinates for the points [tex]P, Q_1, Q_2, P_1, P_2, M[/tex].

    Step 4) As you thought, now use the euclidean distance formula to calculate the length of each of the line segments.

    Hint: For [tex]\overline{PQ_1}[/tex] you should get [tex]||\overline{PQ_1}||^2 = (p_2-p_3)^2\left (p_1^2 + \frac{1}{4}\right )[/tex].

    Notice after you calculate the others that things will start canceling out nicely.

    Step 5) Plug the lengths into your formula to find that they equal 1.

    Good luck.
     
    Last edited: Jul 11, 2009
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