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    (Original post by Zacken)
    It starts from r=0 where as the standard sums start from r=1. So you need to do:

    \displaystyle \sum_{r=0}^{n} \left(r^2 - 2r + 2n + 1\right) = 0^2 - 2(0) + 2n + 1 + \sum_{r=0}^{n} \left(r^2 - 2r + 2n + 1\right)

    Your error was doing (2n+1)(n) + 1 when you should have done (2n+1)(n+1).
    Not 1 to n?
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    (Original post by NotNotBatman)
    Not 1 to n?
    Dumb typo, see edited version.
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    (Original post by Zacken)
    It starts from r=0 where as the standard sums start from r=1. So you need to do:

    \displaystyle \sum_{r=0}^{n} \left(r^2 - 2r + 2n + 1\right) = 0^2 - 2(0) + 2n + 1 + \sum_{r=1}^{n} \left(r^2 - 2r + 2n + 1\right)

    Your error was doing (2n+1)(n) + 1 when you should have done (2n+1)(n+1).
    Ah, that's actually what I had done originally, but couldn't come to an answer. I accidentally treated -2r as 2r , can't believe I lost 5 marks to that.
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    (Original post by NotNotBatman)
    Ah, that's actually what I had done originally, but couldn't come to an answer. I accidentally treated -2r as 2r , can't believe I lost 5 marks to that.
    Ouch!
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    Oh boy oh boy. This is kicking in now. 1 month and 12 days until FP1. Dreading the conics questions, I'm thinking they'll give us a horrid 9 marker or something. Ew!

    Quick question:

    In a question where it gives you a parabola/hyperbola, and tells you the normal to the parabola/hyperbola at P (say P is the general point p(ap^2, 2ap)) meets the parabola again at another point D, how do you go about finding the co-ordinates of this other point Q? Like what is the method and reasoning behind the steps you're taking in order to find the coordinates? And also, where it says 'the tangent at point p and point q meet at the point R, find the coordinates of R' what is the actual method for this? Like for example, why do you find both of the equations of the tangents at P and Q and subtract them rather than adding etc? I can't really explain my question but hopefully you guys can understand what I mean slightly!
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    ^if nobody else has, I'll reply to that in the morning. Just reminding myself.
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    (Original post by iMacJack)
    Oh boy oh boy. This is kicking in now. 1 month and 12 days until FP1. Dreading the conics questions, I'm thinking they'll give us a horrid 9 marker or something. Ew!

    Quick question:

    In a question where it gives you a parabola/hyperbola, and tells you the normal to the parabola/hyperbola at P (say P is the general point p(ap^2, 2ap)) meets the parabola again at another point D, how do you go about finding the co-ordinates of this other point Q? Like what is the method and reasoning behind the steps you're taking in order to find the coordinates? And also, where it says 'the tangent at point p and point q meet at the point R, find the coordinates of R' what is the actual method for this? Like for example, why do you find both of the equations of the tangents at P and Q and subtract them rather than adding etc? I can't really explain my question but hopefully you guys can understand what I mean slightly!
    You use the general equation of the hyperbola at those points and solve simultaneously. For example say
    the tangent equation at P is:  P^2y + x=10p
    and at Q it's q^2y+x = 10q

    Then you want to isolate x and y and then solve simultaneously.
    You would have from the tangent at p,  x=10p - p^2 and from the tangent at Q, x=10q-q^2y

    So,  10q -q^2y = 10p-p^2y

    You can then rearrange for y and get the y coordinates in terms of p and q, then do the same but this time isolating y.

    So,
    -find two tangent equations
    -make x (or y) the subject of both
    -equate these
    -rearrange for y (or x)
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    (Original post by NotNotBatman)
    You use the general equation of the hyperbola at those points and solve simultaneously. For example say
    the tangent equation at P is:  P^2y + x=10p
    and at Q it's q^2y+x = 10q

    Then you want to isolate x and y and then solve simultaneously.
    You would have from the tangent at p,  x=10p - p^2 and from the tangent at Q, x=10q-q^2y

    So,  10q -q^2y = 10p-p^2y

    You can then rearrange for y and get the y coordinates in terms of p and q, then do the same but this time isolating y.

    So,
    -find two tangent equations
    -make x (or y) the subject of both
    -equate these
    -rearrange for y (or x)
    Okay - thank you.
    To make sure I am right, for this one you'd get the equation as above (10q-q^2y=10p-p^2y) what I did was add p^2y to the other side and take away 10q to the other side, this left me with p^2y - q^2y - 10p-10q (isolating the y on one side for factorisation), so then y(p^2-q^2) = 10(p-q)
    y = 10(p-q)/(p-q)(p+q) = 10/(p+q) is that right for the y coordinate? Then plugging back in gives me (10pq)/(p+q) for the X. Could anyone please confirm that my knowledge is correct here? Thank you NotNotBatman appreciated & also just Zacken in general!
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    (Original post by iMacJack)
    Okay - thank you.
    To make sure I am right, for this one you'd get the equation as above (10q-q^2y=10p-p^2y) what I did was add p^2y to the other side and take away 10q to the other side, this left me with p^2y - q^2y - 10p-10q (isolating the y on one side for factorisation), so then y(p^2-q^2) = 10(p-q)
    y = 10(p-q)/(p-q)(p+q) = 10/(p+q) is that right for the y coordinate? Then plugging back in gives me (10pq)/(p+q) for the X. Could anyone please confirm that my knowledge is correct here? Thank you NotNotBatman appreciated & also just Zacken in general!
    Yes, you've got it right.
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    (Original post by NotNotBatman)
    Yes, you've got it right.
    Great, thank you.
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    Could anyone give some help with this question? It's question 14 of CrashMaths (kingaaran)'s coord system pack!

    Part A I need help with!!

    So I've got the normal at point P:

    It works out to be:

    y + px = 2ap + ap^3

    Then we have that y^2 = 4ax, I have then rearranged that so x= y^2/4a, I then subbed this into the normal equation and times'd through by 4a to get rid of the fraction, this gave me

    4ay + py^2 = 8a^2p + 4a^2p^3.

    I'm sure I've gone wrong somewhere, I am beyond confused at how to produce the coordinates wanted in part A

    Many thanks

    kingaaran Zacken aymanzayedmannan
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    (Original post by iMacJack)

    Then we have that y^2 = 4ax, I have then rearranged that so x= y^2/4a, I then subbed this into the normal equation and times'd through by 4a to get rid of the fraction, this gave me

    4ay + py^2 = 8a^2p + 4a^2p^3.

    I'm sure I've gone wrong somewhere, I am beyond confused at how to produce the coordinates wanted in part A



    kingaaran Zacken aymanzayedmannan
    No, you're doing it correctly. Try factorising this - it's the only way. It''s ugly but it can be done.

    You obtained a hidden quadratic in terms of y. My preferred method would be to use the quadratic formula.

    But damn - kingaraan writes some nice questions!
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    (Original post by aymanzayedmannan)
    No, you're doing it correctly. Try factorising this - it's the only way. It''s ugly but it can be done.

    You obtained a hidden quadratic in terms of y. My preferred method would be to use the quadratic formula.

    But damn - kingaraan writes some nice questions!
    He does! They're very fun and challenging I don't know if I'm capable of factorising this o_O
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    (Original post by iMacJack)
    He does! They're very fun and challenging I don't know if I'm capable of factorising this o_O
    Try the quadratic formula!
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    (Original post by aymanzayedmannan)
    Try the quadratic formula!
    What are my a, b and c values :eek:

    I'll be back in a bit - I have my football match now!
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    (Original post by iMacJack)
    What are my a, b and c values :eek:

    I'll be back in a bit - I have my football match now!
    Good luck! I won't ruin it for you just yet
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    (Original post by iMacJack)
    Could anyone give some help with this question? It's question 14 of CrashMaths (kingaaran)'s coord system pack!

    Part A I need help with!!

    So I've got the normal at point P:

    It works out to be:

    y + px = 2ap + ap^3

    Then we have that y^2 = 4ax, I have then rearranged that so x= y^2/4a, I then subbed this into the normal equation and times'd through by 4a to get rid of the fraction, this gave me

    4ay + py^2 = 8a^2p + 4a^2p^3.

    I'm sure I've gone wrong somewhere, I am beyond confused at how to produce the coordinates wanted in part A

    Many thanks

    kingaaran Zacken aymanzayedmannan
    (Original post by aymanzayedmannan)
    No, you're doing it correctly. Try factorising this - it's the only way. It''s ugly but it can be done.

    You obtained a hidden quadratic in terms of y. My preferred method would be to use the quadratic formula.

    But damn - kingaraan writes some nice questions!
    Yeah I think I did this by the quadratic formula too.

    But I did write this question about 8 months ago, so...
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    (Original post by kingaaran)
    Yeah I think I did this by the quadratic formula too.

    But I did write this question about 8 months ago, so...
    Haha completing the square would be tedious too, but I guess it's possible!

    I've done the whole thing, but do you not think that part c is FP3 material? I don't think loci problems are in FP1.
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    (Original post by aymanzayedmannan)
    Haha completing the square would be tedious too, but I guess it's possible!

    I've done the whole thing, but do you not think that part c is FP3 material? I don't think loci problems are in FP1.
    Yeah potentially. In my papers, I put strictly on-spec material, but for my worksheets, well, we can have some fun, can't we
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    (Original post by kingaaran)
    Yeah potentially. In my papers, I put strictly on-spec material, but for my worksheets, well, we can have some fun, can't we
    Perfect - gonna scour them for similar stuff.
 
 
 
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