Answer
$y=a-a\cdot cos(\frac{x+\sqrt {2ay-y^2}}{a})$
Work Step by Step
Step 1. Recall the equations obtained in Example 6 as $x=a(\theta-sin\theta)$, $y=a(1-cos\theta)$.
Step 2. The first equation gives $\theta=\frac{x}{a}+sin\theta$
Step 3. With the second equation, we have $cos\theta=1-\frac{y}{a}=\frac{a-y}{a}$ and $\theta=cos^{-1}(\frac{a-y}{a})$
Step 4. Within $0\leq\theta\leq \pi$, we have $sin\theta=\sqrt {1-(\frac{a-y}{a})^2}=\frac{\sqrt {2ay-y^2}}{a}$
Step 5. Combine the results above, we get: $cos^{-1}(\frac{a-y}{a})=\frac{x}{a}+\frac{\sqrt {2ay-y^2}}{a}$
Step 6. Take cosine on both sides of above to get $\frac{a-y}{a}=cos(\frac{x+\sqrt {2ay-y^2}}{a}$) or
$y=a-a\cdot cos(\frac{x+\sqrt {2ay-y^2}}{a})$
