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    i understand part one because someone has already helped me with it but part two...
    i dont know how to get the coordinates for point Q and yes i know that y=0 on it but i dont know how to get the x coordinate.

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    (Original post by cooldudeman)
    i understand part one because someone has already helped me with it but part two...
    i dont know how to get the coordinates for point Q and yes i know that y=0 on it but i dont know how to get the x coordinate.

    use the info from part (i) to form an equation for the normal using
    y - y1 = m(x -x1)
    then solve the resulting equation to find the value of x that gives y=0
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    (Original post by cooldudeman)
    i understand part one because someone has already helped me with it but part two...
    i dont know how to get the coordinates for point Q and yes i know that y=0 on it but i dont know how to get the x coordinate.

    y = -2x/3

    parallel to this line, therefore the line is:

    y = -2x/3 + c, where c is a real constant

    i dont remember whether C1 has differentiation, but if so:

    dy/dx = -1/(2sqrt(x))

    if x =4, y = ? for -2x/3 and for -1/(2sqrt(x))


    these lines should then be shown to be parallel
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    You need to find the equation of the normal then let y = 0 to find where it intersects with the x axis.
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      (Original post by Femto)
      You need to find the equation of the normal then let y = 0 to find where it intersects with the x axis.
      For the equation of the normal, I got y = (1/4)kx + 3k

      Wtf?
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        I've got:

        y=-2/3 x
        as it is a normal it will be
        y=-1/(-2/3) x
        this equals to
        y=-3/2 x
        i.e.
        y=-3x/2

        what do I do next?
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        (Original post by im so academic)
        For the equation of the normal, I got y = (1/4)kx + 3k

        Wtf?
        Yes because the preceding part of the question asks you to find k which I believe is 6 in this case.
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        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.
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        (Original post by Femto)
        Yes because the preceding part of the question asks you to find k which I believe is 6 in this case.

        (Original post by im so academic)
        I've got:

        y=-2/3 x
        as it is a normal it will be
        y=-1/(-2/3) x
        this equals to
        y=-3/2 x
        i.e.
        y=-3x/2

        what do I do next?

        (Original post by cooldudeman)
        i understand part one because someone has already helped me with it but part two...
        i dont know how to get the coordinates for point Q and yes i know that y=0 on it but i dont know how to get the x coordinate.

        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.
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        (Original post by im so academic)
        I've got:

        y=-2/3 x
        as it is a normal it will be
        y=-1/(-2/3) x
        this equals to
        y=-3/2 x
        i.e.
        y=-3x/2

        what do I do next?
        Which part are you asking about?
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          (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.
          Found the gradient at P = 1/4k
          Therefore the normal will be 4k????
          But the other line has a gradient of -3/2
          How is it 6?
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            (Original post by Femto)
            Which part are you asking about?
            Trying to work out the value of k.
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            (Original post by im so academic)
            Found the gradient at P = 1/4k
            Therefore the normal will be 4k????
            But the other line has a gradient of -3/2
            How is it 6?
            Well y = k\sqrt x right? If the normal is parallel to 2x + 3y = 0 then the tangent gradient is needed; this will be \frac{3}{2}

            We are given that x = 4 at point P so if we differentiate y = k \sqrt x:

            \frac{dy}{dx} = \frac{k}{2\sqrt x}

            Set \frac{dy}{dx} = \frac{3}{2} and x = 4

            We get:

            \frac{k}{4} = \frac{3}{2}

            \rightarrow k = \frac{12}{2}

            Which means k = 6

            (I've only explained the first part since OP said he knows how to do it.)
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              OK this is what I've done now.

              Equation of gradient at normal is y = 4kx
              Equation of the other line is y = -2/3x

              Equalise:

              4kx = -2/3x

              Cancel out x's:

              4k = -2/3

              Rearrange for K:

              k = -1/6

              What?
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              (Original post by im so academic)
              Found the gradient at P = 1/4k
              Therefore the normal will be 4k????
              The gradient of the normal at P is not 1/4k. I'm not sure what you mean by the normal being '4k'?...

              y=k\sqrt{x} \implies \frac{dy}{dx}=\frac{k}{2\sqrt x}
              When x=4, \frac{dy}{dx}=\frac{k}{4}.

              This is the gradient of the TANGENT at P. You know that tangents and normals at a point are perpendicular to each other. From this, what can you tell me about the gradient of the normal at that point?

              You know that this normal is parallel to 2x+3y=0 i.e. it has the same gradient. What is the gradient of this line? How does it relate to the gradient above?
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                Wait why is the tangent gradient needed? Doesn't normal = perpendicular?

                Anyways, surely if you rearrange that equation for y you get -2/3?
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                  (Original post by Farhan.Hanif93)
                  y=k\sqrt{x} \implies \frac{dy}{dx}=\frac{k}{2\sqrt x}
                  How did you get k over 2 root x?
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                  (Original post by Femto)
                  Well y = k\sqrt x right? If the normal is parallel to 2x + 3y = 0 then the tangent gradient is needed; this will be \frac{3}{2}

                  We are given the x coordinate 4 so if we differentiate y = k \sqrt x:

                  \frac{dy}{dx} = \frac{k}{2\sqrt x}

                  Set \frac{dy}{dx} = \frac{3}{2} and x = 4

                  We get:

                  \frac{k}{4} = \frac{3}{2}

                  \rightarrow k = \frac{12}{2}

                  Which means k = 6

                  (I've only explained the first part since OP said he knows how to do it.)
                  Even though this problem works by considering the equation of the tangent and the equation of the perpendicular to 2x+3y=0 through the origin, as you have shown. This is not the intended method and may lose you marks in the exam. You need to consider the NORMAL at P and the line 2x+3y=0.

                  I think what you've posted may confuse a few as it's the roundabout way of doing it.
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                  (Original post by im so academic)
                  How did you get k over 2 root x?
                  k\sqrt x = k \times x^{\frac{1}{2}}
                  Recall that if y=ax^n then \frac{dy}{dx}=anx^{n-1} i.e. multiply by the original power and then lower the power by 1.
                 
                 
                 
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