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Hi i have answered parts 1 and 2 of this question am stuck on part 3. 1 A curve has a parametric equations x=t^2 y=4t Find the equation of the normal to this curve at (t^2,4t) (i have solved this using parametric differentiation and getting the equation as tx+2y-8t-t^3 using y-y1=m(x-x1) we know that y=4t and x=t^2 and dy/dx is what we have solved using parametric differentiation) 2.find the coordinate of the points where the normal cuts the coordinates axes. setting y=0 x= 8+t^2) 3.find in terms of t, the area of the triangle enclosed by the normal and the axes.(where am stuck)
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the answer to part 3 should be (0,0.5t{8+t^2})
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