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    This years paper was generally a nice paper, we had an easy matrices proof by induction and we barely scratched the surface of summing a series. I got to the final question with 30 minutes to spare.

    However, Question 8 was a joke, here is the question:

    y^2= 12x, the point P is at point P(3p^2, 6p)

    S is the focus

    a) Show that SP is 3(1+p^2)

    Q= (3q^2, 6q) p=/=q

    b) The tangent at point P on the curve and the tangent of point Q on the curve meet at point R, find the coordinates of the point R

    c) Show that RS= SP.SQ
    ________________________________ ________________________

    Part a) was fine, if you didn't know you had to Pythagoras the change in x and change in y and simplify.

    Part b)

    I was stuck on this question, like no method I tried seemed to work right, I would always get p=q.

    First, I found the equation of the tangent: y = 1/p x + 3p

    Then subbed in q (Coords were the same algebraicly) y= 1/q x +3q

    then I'm stumped.

    Normally you would sub in the values of the point into the y and x variables and solve a quadratic equation, this time, there is no such luck.

    I tried to sub in the y value and eliminate a variable: Nope.

    I tried to sub in an x value and eliminate a variable: Nope

    Were you supposed to sub in x = y^2 /12 ?, I'm honestly concerned that I'm just forgetting something.

    Edit: EDEXCEL ARE NOT NICE PEOPLE.

    You basically had to equate one another and get x/y on its own.

    I could have sworn the question didn't specify for an answer in terms of p and q, I was trying to find their values....

    Thanks for the answers guys!
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    (Original post by XI Ki11JoY IX)
    This years paper was generally a nice paper, we had an easy matrices proof by induction and we barely scratched the surface of summing a series. I got to the final question with 30 minutes to spare.

    However, Question 8 was a joke, here is the question:

    y^2= 12x, the point P is at point P(3p^2, 6p)

    S is the focus

    a) Show that SP is 3(1+p^2)

    Q= (3q^2, 6q) p=/=q

    b) The tangent at point P on the curve and the tangent of point Q on the curve meet at point R, find the coordinates of the point R

    c) Show that RS= SP.SQ
    ________________________________ ________________________

    Part a) was fine, if you didn't know you had to Pythagoras the change in x and change in y and simplify.

    Part b)

    I was stuck on this question, like no method I tried seemed to work right, I would always get p=q.

    First, I found the equation of the tangent: y = 1/p x + 3p

    Then subbed in q (Coords were the same algebraicly) y= 1/q x +3q

    then I'm stumped.

    Normally you would sub in the values of the point into the y and x variables and solve a quadratic equation, this time, there is no such luck.

    I tried to sub in the y value and eliminate a variable: Nope.

    I tried to sub in an x value and eliminate a variable: Nope

    Were you supposed to sub in x = y^2 /12 ?, I'm honestly concerned that I'm just forgetting something.
    i got same for equations. then i equated p equation with q equation as got we both got it in terms of y =. then solved to get x co ordinates ?
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    If y=(1/p)X+3p and y=(1/q)X+3q, then just equate them to each other, so (1/p)X+3p=(1/q)X+3q, the solve for X. Afterwards, put the tangent equations as x in terms of y (e.g. x=py-3p^2) and equate them to each other again to find the value of y.
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    The tangent lines are py=x+3p^2 and qy=x+3q^2. Make y the subject of each and equate both of them to get x=3pq. Then do the same for x and you get y=3(p+q).
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    Part (a) I don't think required any real calculation - all did was to add 3 to the x coordinate of P (to find distance of P from the directrix). This gives you 3p^2 + 3 [aka 3(p^2+1)], which we know from the definition of a parabola to be the same as the distance SP.
    Regretting not putting more down.
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    (Original post by Don John)
    The tangent lines are py=x+3p^2 and qy=x+3q^2. Make y the subject of each and equate both of them to get x=3pq. Then do the same for x and you get y=3(p+q).
    i left the equations as the original post. therfore i got x being -3(q-p)^2 ?
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    is this edexcel gcse maths
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    (Original post by Maths degree)
    i left the equations as the original post. therfore i got x being -3(q-p)^2 ?
    \frac{1}{q}x+3q=\frac{1}{p}x+3p
    x\left(\frac{1}{q}-\frac{1}{p}\right)=3(p-q)
    x=\frac{3(p-q)}{\left(\frac{1}{q}-\frac{1}{p}\right)} = \frac{3(p-q)}{\left(\frac{p-q}{pq}\right)} = 3pq
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    (Original post by Don John)
    \frac{1}{q}x+3q=\frac{1}{p}x+3p
    x\left(\frac{1}{q}-\frac{1}{p}\right)=3(p-q)
    x=\frac{3(p-q)}{\left(\frac{1}{q}-\frac{1}{p}\right)} = \frac{3(p-q)}{\left(\frac{p-q}{pq}\right)} = 3pq
    Literally realised that before going to sleep last night lol, I feel stupid.
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    (Original post by XI Ki11JoY IX)
    Literally realised that before going to sleep last night lol, I feel stupid.
    I think anyone that thought over any questions they struggled with (uauslly 8c) realised how to do it before going to sleep... I did... shame it was a bit late So don't feel stupid. It happens to everyone.
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    (Original post by Tiva4Eva)
    I think anyone that thought over any questions they struggled with (uauslly 8c) realised how to do it before going to sleep... I did... shame it was a bit late So don't feel stupid. It happens to everyone.
    I can't remember 8c, never mind though, I'm still looking for a high B
 
 
 

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