A cycloid is generated by rolling a circle of radius
A cycloid is generated by rolling a circle of radius $a$ along a straight line. Prove that the normal to the cycloid at any point $M$ passes through the current contact point $Q$ of the rolling circle with the line.
Explanation of Method
A classical fact about the cycloid, proved by matching two slopes.
Step 1: Setup
With the base line as the $x$-axis and rotation angle $t$, the centre of the rolling circle is $C=(at,a)$, the traced point is
\[ M = \bigl(a(t-\sin t),\; a(1-\cos t)\bigr), \tag{1} \]
and the current contact point, directly below $C$, is
\[ Q = (at,\,0). \tag{2} \]
Step 2: Slope of the tangent and normal at $M$
As in Problem 532, $k_{t}=\cot\tfrac t2$, so
\[ k_{n} = -\tan\frac t2. \tag{3} \]
Step 3: Slope of $MQ$
\[ k_{MQ} = \frac{0-a(1-\cos t)}{at-a(t-\sin t)} = -\frac{1-\cos t}{\sin t} = -\frac{2\sin^{2}\tfrac t2}{2\sin\tfrac t2\cos\tfrac t2} = -\tan\frac t2. \tag{4} \]
Step 4: Conclusion
Since $k_{n}=k_{MQ}=-\tan\tfrac t2$ and both lines pass through $M$, the normal at $M$ is the line $MQ$; it therefore passes through $Q=(at,0)$. $\blacksquare$
Step 5: Verification
\[t=0.9,\ a=1\text{: } k_n=-\tan0.45\approx-0.48306,\ k_{MQ}\approx-0.48306\ \checkmark\]
Result
Final answer:
\[\boxed{\,k_{n} = k_{MQ} = -\tan\frac t2\,}\]
so the normal to the cycloid at every point $M$ passes through the current contact point $Q=(at,0)$ of the rolling circle.
Remark
Remark. Geometrically, $M$ lies on the rolling circle and $Q$ is its point of contact, so $MQ$ is a chord of that circle. Since the diameter through $Q$ is vertical (perpendicular to the base line), the tangent–chord angle at $M$ forces the chord $MQ$ itself to be the normal direction to the cycloid at $M$ — the algebraic identity above is the coordinate form of this fact.