Consider the epicycloid x=a [(1+ lambda) cos t- lambda

Consider the epicycloid $x=a\!\left[(1+\lambda)\cos t-\lambda\cos\mu t\right],\ y=a\!\left[(1+\lambda)\sin t-\lambda\sin\mu t\right]$, $\mu=1+\tfrac1\lambda$, $\lambda>0$, and the circle $x^{2}+y^{2}=a^{2}$. Prove that the normal to the epicycloid at the point with parameter $t$ intersects the circle at the point $Q=(a\cos t,\,a\sin t)$.

Step 1: Derivatives

Using $\lambda\mu=\lambda+1$: \[\dot x = a(1+\lambda)\bigl(\sin\mu t-\sin t\bigr), \qquad \dot y = a(1+\lambda)\bigl(\cos t-\cos\mu t\bigr). \tag{1}\]

Step 2: Slope of the tangent

Let $\alpha=\tfrac{(\mu+1)t}{2}$, $\beta=\tfrac{(\mu-1)t}{2}$; by sum-to-product, $\cos t-\cos\mu t=2\sin\alpha\sin\beta$ and $\sin\mu t-\sin t=2\cos\alpha\sin\beta$, so

\[ k_{t} = \frac{\dot y}{\dot x} = \frac{2\sin\alpha\sin\beta}{2\cos\alpha\sin\beta} = \tan\alpha. \tag{2} \]

Step 3: Slope of the normal, and of $MQ$

$k_{n}=-\cot\alpha$. Writing $M=(x_0,y_0)$ for the point at parameter $t$ and computing $a\cos t-x_0$ and $a\sin t - y_0$ directly from the parametrisation (each collapses to a factor of $a\lambda$ times a sum-to-product expression):

\[ a\cos t - x_{0} = a\lambda(\cos\mu t-\cos t), \qquad a\sin t - y_{0} = a\lambda(\sin\mu t-\sin t). \tag{3} \]

So, applying the same sum-to-product identities to this ratio, \[\frac{a\sin t-y_{0}}{a\cos t-x_{0}} = \frac{\sin\mu t-\sin t}{\cos\mu t-\cos t} = \frac{2\cos\alpha\sin\beta}{-2\sin\alpha\sin\beta} = -\cot\alpha = k_{n}. \tag{4}\]

Step 4: Conclusion

Equation (4) says exactly that $Q=(a\cos t,a\sin t)$ lies on the normal line at $M$. And since $(a\cos t)^{2}+(a\sin t)^{2}=a^{2}$, $Q$ also lies on the circle. $\blacksquare$

Step 5: Verification

\[(a,\lambda,t)=(2,1.5,0.7)\text{: } Q \text{ satisfies the normal equation (residual} \sim10^{-10}\text{)}\ \checkmark\]

\[(a,\lambda,t)=(1,0.6,1.3),\ (3,2.2,2.1)\text{: same check}\ \checkmark\]