Problem

Figure 4 shows a sketch of the closed curve with equation

$$(x+y)^3+10y^2=108x$$

(a) Show that

$$\frac{dy}{dx}=\frac{108-3(x+y)^2}{20y+3(x+y)^2}$$

The curve is used to model the shape of a cycle track with both $x$ and $y$ measured in km.

The points $P$ and $Q$ represent points that are furthest north and furthest south of the origin $O$, as shown in Figure 4.

Using the result given in part (a),

(b) find how far the point $Q$ is south of $O$. Give your answer to the nearest $100$ m.