Originally Posted by

**dojo** A curve is defined by the parametric equations *x* = *a*cos^2 *t*, *y* = *a*sin^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.

OK.

(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???

Thanks