I have no problems with drawing the graph. I got that function is symmetrical about the x axis, roots at $\displaystyle ( 0 , 0 )$ and $\displaystyle (3a , 0 )$ and a turning point $\displaystyle \left( a , \pm \frac{2}{3} a \right)$ you have a closed loop between 0 to 3a and then y just increases with x increasing.Quote:

A curve has equation $\displaystyle 9ay^2= x(3a-x)^2$

where a is a positive constant

a) Sketch a graph of the curve and showing stationary points and intersection with the axes

b) show that the arc length of C in the first quadrant form the point $\displaystyle x = 4a$ up to $\displaystyle x = 9a$ is given by

$\displaystyle \frac{1}{2} \int_{4a}^{9a}{\left[ \left( \frac{a}{x} \right )^{\frac{1}{2}} + \left( \frac{x}{a} \right )^{\frac{1}{2}} \right] dx}$

My only problem is the arc length part. I could do with a pointer.

Thanks Bobak