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Parametric parabolic loci

  1. Oct 3, 2008 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    The points [tex]P(2ap, ap^{2})[/tex] and [tex]Q(2aq, aq^{2})[/tex] lie on the parabola [tex]x^{2} = 4ay[/tex].
    The equation of the normal to the parabola at P is [tex]x + py = 2ap + ap^{3}[/tex] and
    the equation of the normal at Q is [tex]x + qy = 2aq + aq^{3}[/tex]. These normals intersect at R. Find the locus of R if PQ is a focal chord.


    2. Relevant equations
    equation of line:[tex]\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

    [tex]\frac{dy}{dx}=p=q[/tex], p & q are parameters on the parabola [tex]x^{2}=4ay[/tex]

    focal chord (chord passing through focus [0,a]) is given by: [tex]y=\frac{1}{2}(p+q)x-apq[/tex]
    if focal chord, pq=-1


    3. The attempt at a solution
    The coordinates of R is [tex] ( –apq[p + q] , a[p^{2}+pq+q^{2}+2] ) [/tex] and since PQ is a focal chord, pq=-1, therefore this simplifies to [tex] ( -a[p+q] , a[p^{2}+q^{2}+1] ) [/tex]
    From here I am completely stumped on what I need to do. I'm even unsure if the coordinates of R is necessary in this question.
    Any ideas of suggestions for how I can begin to approach this question?
     
    Last edited: Oct 3, 2008
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  3. Oct 4, 2008 #2

    tiny-tim

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    … what would Euclid have done?

    Hi Mentallic! :smile:

    Are you allowed to use ancient Greek geometry?

    Draw the figure nice and large, and draw an extra couple of lines parallel to the axis of the parabola, and work out the angles (and remember why it's called the focus! :wink:)
     
  4. Oct 5, 2008 #3

    Mentallic

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    ...I honestly have no idea.

    We have never used Euclidean geometry as a class. The only locus we have done so far is using the ideas that a curve is a certain distance from a point or line.

    e.g. the locus of the curve that is always 4 units from the origin. By using the distance formula: [tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=4[/tex] simplified to [tex]x^{2}+y^{2}}=4[/tex]

    or for e.g.2 the locus of the curve that is always equal distance between (-2,2) and (2,-2) by using the distance formula and simplifying, becomes [tex]y=x[/tex].

    I don't see a possible way of applying this idea to find the locus of R. However, since pq=-1, this means that the gradients of the normal coming from P and the normal from Q are perpendicular to each other. Maybe I can find the locus by finding the curve which satisfies the two normals to the parabola that are perpendicular to each other?
     
  5. Oct 5, 2008 #4

    tiny-tim

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    Hi Mentallic! :smile:

    Yes, that's the important point … that the two normals are perpendicular …

    you could also have proved this by using the fact that lines PP' and QQ' parallel to the y-axis hit the parabola and are "focussed" onto the focus … so each normal bisects the angle between QPP' and PQQ' … so the angle between the normals is half of QPP' + PQQ', which is 180º. :smile:
    Ok, back to reality :biggrin: … so the locus has x = a(p+q), y = a(p2 + 1 + q2).

    You need an equation relating x and y.

    Hint: what is x2? :wink:
     
  6. Oct 5, 2008 #5

    Mentallic

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    I am very bad at geometry (even worse than my locus skills :yuck:) so I don't even know what the P' and Q' points are.
    Using this big hint :smile:, it seems I need to manipulate the x and y coordinates to become equal.
    The coordinates of R is … [tex] ( -a[p+q] , a[p^{2}+q^{2}+1] ) [/tex]
    So,
    [tex]y=a(p^{2}+q^{2}+1)[/tex]
    [tex]x=-a(p+q)[/tex]
    [tex]x^{2}=a^{2}(p^{2}+2pq+q^{2})[/tex]
    [tex]\frac{x^{2}}{a}=a(p^{2}+q^{2}+1)+a(2pq-1)[/tex]
    [tex]\frac{x^{2}}{a}-a(2pq-1)=y[/tex]
    but pq=-1

    Therefore, the locus is ~
    [tex]y=\frac{x^{2}}{a}+3a[/tex]

    I tested the locus and it is correct. Thanks for showing me another method to find loci, tiny-tim. If it weren't for the fact that I knew the locus would be parabolic and being given the hint you gave me, I probably wouldn't have noticed that I needed to find x2 to take that step closer to make it equal to y. Is the best solution just to keep my mind open and hope I realize the similarities, or are there little things to look out for when relating the x and y terms?
     
  7. Oct 5, 2008 #6

    tiny-tim

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    P' and Q' are any points on the same "vertical" line as P and Q, respectively, so that PP' and QQ' are parallel to the axis of the parabola.

    All lines parallel to the axis are reflected onto the focus … that's why reflecting telescopes have parabolic mirrors … see http://en.wikipedia.org/wiki/Parabolic_reflector
    Yes … basically, it's just looking for a pattern …

    though in this case you might have got there more quickly if you'd realised the importance of pq = -1 as soon as you found it! :wink:
     
  8. Oct 5, 2008 #7

    Mentallic

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    Ahh yes, we have used and proven these ideas for car headlights and face mirrors.

    Oh it wasn't hard once I knew I had to find x2. The rearranging I am ok with, knowing that I need to square/cube/root I am not so alright with. Something I have to teach myself.
     
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