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Math Help - Loci...?

  1. #1
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    Loci...?

    The straight line joining the points P(2p,p) and Q(2q,q) passes through the point (0,1)

    a) i) Show that pq = -1

    ii) Deduce the coordinates of R, the midpoint of PQ, are

    (p - 1/p , 1/2(p + 1/p))


    b) Find the cartesian equation of the locus of R as p varies, giving your answer in the form y=f(x)



    OK i have attempted a lame start.

    x=2p y=p

    p=x/2 y=(x/2)

    and now i'm stuck
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  2. #2
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    Hi

    Let A(0,1)
    We know that \overrightarrow{AP} and \overrightarrow{QP} are colinear

    \overrightarrow{AP} coordinates are (2p,p-1)
    \overrightarrow{QP} coordinates are (2p-2q,p-q)

    Therefore 2p(p-q)-(2p-2q)(p-1)=0

    Spoiler:

    -2pq+2qp+2p-2q=0
    pq(-q+p)+(p-q)=0
    (p-q)(pq+1)=0
    Since p and q are different => pq=-1
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  3. #3
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    Quote Originally Posted by djmccabie View Post

    The straight line joining the points P(2p,p) and Q(2q,q) passes through the point (0,1)

    a) i) Show that pq = -1
    assuming that P and Q are distinct, we must have p \neq q \neq 0. now look at the slopes: \frac{p^2 - 1}{2p} = \frac{q^2 - 1}{2q}, which after simplifying gives you: (pq + 1)(q-p)=0 and so pq=-1.


    ii) Deduce the coordinates of R, the midpoint of PQ, are (p - 1/p , 1/2(p + 1/p))
    use the formula you have for the midpoint coordinates and then using the first part of your problem put q=\frac{-1}{p}.


    b) Find the cartesian equation of the locus of R as p varies, giving your answer in the form y=f(x)

    OK i have attempted a lame start.

    x=2p y=p

    p=x/2 y=(x/2)

    and now i'm stuck
    you don't need to do anything else! your answer, y=\frac{x^2}{4}, is corretc.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    assuming that P and Q are distinct, we must have p \neq q \neq 0. now look at the slopes: \frac{p^2 - 1}{2p} = \frac{q^2 - 1}{2q}, which after simplifying gives you: (pq + 1)(q-p)=0 and so pq=-1.
    Thanks very much for your help I'm still a bit confused. How did you deduce (p^2-1)/2p = (q^2-1)/2q

    Also i can't see how it simplifies to that could you add each step please its killing me
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  5. #5
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    Quote Originally Posted by djmccabie View Post
    Thanks very much for your help I'm still a bit confused. How did you deduce (p^2-1)/2p = (q^2-1)/2q

    Also i can't see how it simplifies to that could you add each step please its killing me
    We know that \overrightarrow{AP} and \overrightarrow{AQ} are colinear

    \overrightarrow{AP} coordinates are (2p,p-1)
    \overrightarrow{AQ} coordinates are (2q,q-1)

    Therefore (p-1)/2p = (q-1)/2q
    q(p-1) = p(q-1)
    pq-q-pq+p=0
    pq(p-q)+(p-q)=0
    (p-q)(pq+1)=0
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  6. #6
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    Quote Originally Posted by running-gag View Post
    \overrightarrow{AP} coordinates are (2p,p-1)
    \overrightarrow{AQ} coordinates are (2q,q-1)
    Hi thanks a lot for your help. I'm still a bit unsure how you derive this, I think i may just be having a mental block!

    (2p,p-1) and (2q,q-1)

    Now I realise this comes from the point A(0,1) but im not sure how



    Quote Originally Posted by running-gag View Post
    Therefore (p-1)/2p = (q-1)/2q
    I'm also unsure how this is derived :/

    sorry if this is really simple math think i may need to take a break
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  7. #7
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    If A(x_A,y_A) and B(x_B,y_B) then \overrightarrow{AB}(x_B-x_A,y_B-y_A)

    Here A(0,1) and P(2p,p^2) therefore \overrightarrow{AP}(2p,p^2-1)

    Now if \vec{u}(u_x,u_y) and \vec{v}(v_x,v_y) are colinear then u_x\:v_y - u_y\:v_x = 0

    You can also use the slope \frac{u_y}{u_x} = \frac{v_y}{v_x} provided that u_x \mbox{and} v_x \neq 0
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  8. #8
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    Quote Originally Posted by running-gag View Post
    You can also use the slope \frac{u_y}{u_x} = \frac{v_y}{v_x} provided that u_x \mbox{and} v_x \neq 0

    Is this also only if colinear?
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  9. #9
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    Yes

    u_x\:v_y - u_y\:v_x = 0 \Rightarrow u_x\:v_y = u_y\:v_x and dividing by u_x\:v_x leads to \frac{v_y}{v_x} = \frac{u_y}{u_x}
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  10. #10
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    Thanks dude!! real help
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