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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- Thread Starter
- 11-05-2015 22:05
- 12-05-2015 11:55
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