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    umm so many people replied... i dont know which is the correct explanation for the second part.
    the first part of the question is like this (as i was already explained for this one):
    point P has x coordinate 4 on the curve y= k(rootx) <--- turn this into y=kx^-1 for differentiation
    dy/dx= k x 1/2 x (x)^-1/2 then if we sub the x=4 in it, it would be:
    dy/dx= 1/4(k)
    ----------
    then we have got the equation, 2x+3y=0 which is parallel to the normal of P. and if we rearrange this into y=mx+c, we get:
    y= -2x/3 we can clearly see from this that the gradient of the normal of point P is -2/3 (since is it PARALLEL, it is the same as the normal on point P)
    now if we do the negative reciprecol thing, we can get the gradient of the tangent of P because differentiation finds the gradient of the tangent only. -2/3 --> 3/2. then we sub the 3/2 into the dy/dx from above.

    3/2= 1/4(k) then we can find k which must be 6
    k=6

    btw the actual coordinates of point Q on the second part are, (0,22) but i dont have any idea how they got this. it doesnt say on the mark scheme...
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    (Original post by Femto)
    Ok if k = 6 then we can assume the following:

    y = 6\sqrt x

    Since the line y = -\frac{2}{3}x is parallel to the normal at the point which x = 4 then the equation of the normal can be worked out since we have the point P (4, 12) and a gradient of -\frac{2}{3}.

    I think you can carry on from there.
    omg you're a genius!
    i just tried what you said and it works. when it said, 'This normal' i had no idea that it meant the one for point P! the wording is so weird...
    also, how do i know that it is a right angled triangle? because the formula, 1/2 x b x h, is used.
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    (Original post by cooldudeman)
    btw the actual coordinates of point Q on the second part are, (0,22) but i dont have any idea how they got this. it doesnt say on the mark scheme...
    Read the following:
    (Original post by Farhan.Hanif93)
    You don't need to find the equation of the normal. You only need to find the gradient of the normal at x=4 in terms of k using differentiation. Then note that parallel lines have equal gradients. You know what the gradient of 2x+3y=0 is so equate the two expressions and solve for k.
    k is 6.

    for (ii) find the x-coordinate of Q and the y-coordinate of P and then use A=\frac{1}{2}base\times height.
    Find the value of y=6\sqrt x at P i.e. when x=4, what is y?
    You should also notice that the normal passes through this point.
    You now want to find the equation of the normal. If you're familiar with the method of using the equation (y-y_1)=m(x-x_1) where (x_1,y_1) is the coordinates of P to find the equation of the normal, go for it.

    For the triangle, it may help to draw a sketch of the lines which make up the sides of the triangle. The only lengths that matter are the base of the triangle i.e. the distance from the origin to Q; and the height of the triangle i.e. the distance from P to Q.
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    (Original post by im so academic)
    There is in Edexcel and AQA, not sure about OCR specifications though.
    This is an OCR question
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      (Original post by Femto)
      This is an OCR question
      Oh thank God, no wonder it looked a bit tricky to me. It is true, there are differences between exam boards.
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      (Original post by im so academic)
      Oh thank God, no wonder it looked a bit tricky to me. It is true, there are differences between exam boards.
      There isn't much of a difference in difficulty across the exams boards. OCR MEI has a reputation for being difficult but this question is from the normal OCR specification (I seem to remember it from C1).
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      i dont know.
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      I'm on the second part now. I got 2x+3y=44 but the markscheme says its at points (22,0)

      How do i get to those points from what i have worked out (providing i have done it right :/ )
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      (Original post by Robpattinsonxxx)
      I'm on the second part now. I got 2x+3y=44 but the markscheme says its at points (22,0)

      How do i get to those points from what i have worked out (providing i have done it right :/ )
      Well, the normal will intersect with the x axis when y = 0 therefore:

      2x + 3(0) = 44

      ...carry on
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      x=22.

      Thanks so much, genious I've seen you around a lot! Thanks x
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      I did this.

      We know that the normal grad= 3/2

      and the co-ordinates are (0,0) so eqn= 2y=3x

      stick 4 into 3x -> 2y=3(4), therefore y=6 and so the value of k =6.
     
     
     
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