A curve is defined by the parametric equations x = acos^2 t, y = asin^3 t
for range 0<= t <= pi/2 where a is a positive constant.
(i) Find and simplify an expresion for dy/dx in terms of t
(ii) Find the equation of the noraml at the point on the curve where
t = pi/6
(ii) If the normal cuts the the axis at point M and N, find the area of the triangle of the triangle OMN, in terms of a, where O is the origin.
(i)after finding dx/dt and dy/dt I simplfy to get -3/2(sin t)
(ii) The grad. of tangent is -3/4 when t = pi/6. Therefore the normal will be 4/3
(iii) I'm lost with this part...
Am I correct up to (iii) and can anyone help???