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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
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    (Original post by skate-elz)
    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 need to find the equation of the normal line, so you can see where this line crosses the x axis (point Q).
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    i.e the actual normal. 2x+3y=0 is only parallel to the normal at P. Once you know the specific normal at P, you can set y=0 to find the x coordinate at Q. You now have the three corners of your triangle.
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    Thankyou both!!
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    (Original post by skate-elz)
    Thankyou both!!
    No problem let us know if you get stuck.
 
 
 
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