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    The question is The point P on the curve y = k√x has x-coordinate 4. The normal to the curve at P is parallel tothe line 2x + 3y = 0.

    I did dy/dx of k √ x and equated it to 3/2 as this is the gradient of the tangent at P and solved k to = 6.

    The next question is (ii) This normal meets the x-axis at the point Q. Calculate the area of the triangle OPQ, where O isthe point (0, 0).

    I have worked out the coordinate P (4,12) and O = (0, 0) and i assumed the Y coordinate of Q to be 0 but i need to know how to find x coordinate. Q= (x, 0)
    If somebody could explain to me how to do this so i can apply it to this question and to future questions it would be highly appreciated!! Thankyou

    You can work out the equation of the normal as you are already given the gradient and a point it passes through.
    Since it is parallel to 2x+3y=0 you can sub in P(4,12) to obtain the normal. 2*4+3*12=44, so normal is 2x+3y=44.
    It meets x-axis when y = 0 so 2x=44 and therefore x=22.
    Since both O and Q are on the x-axis the vertical height of your triangle is just the y coordinate of P.
    Thus the area of the triangle is (12*22)/2 = 132 sq units.

    Hope that makes sense and offers some help
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Updated: May 12, 2015


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