Tangent to Cycloid passes through Top of Generating Circle

Theorem

Let $C$ be a cycloid generated by the equations:

$x = a \paren {\theta - \sin \theta}$

$y = a \paren {1 - \cos \theta}$

Then the tangent to $C$ at a point $P$ on $C$ passes through the top of the generating circle of $C$.

From Tangent to Cycloid, the equation for the tangent to $C$ at a point $P = \tuple {x, y}$ is given by:

$(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$

From Equation of Cycloid, the point at the top of the generating circle of $C$ has coordinates $\tuple {2 a, a \theta}$.

Substituting $x = 2 a$ in $(1)$:

$\ds y - a \paren {1 - \cos \theta}$

$=$

$\ds \dfrac {\sin \theta} {1 - \cos \theta} \paren {a \theta - a \theta + a \sin \theta}$

$\ds \leadsto \ \ $

$\ds y$

$=$

$\ds a \paren {1 - \cos \theta} + \dfrac {\sin \theta} {1 - \cos \theta} a \sin \theta$

$\ds $

$=$

$\ds \frac {a \paren {1 - \cos \theta}^2 + a \sin^2 \theta} {1 - \cos \theta}$

$\ds $

$=$

$\ds \frac {a \paren {1 - 2 \cos \theta + \cos^2 \theta} + a \sin^2 \theta} {1 - \cos \theta}$

$\ds $

$=$

$\ds \frac {a \paren {1 - 2 \cos \theta + 1} } {1 - \cos \theta}$

$\ds $

$=$

$\ds \frac {a \paren {2 - 2 \cos \theta} } {1 - \cos \theta}$

$\ds $

$=$

$\ds 2 a$

That is, the tangent to $C$ passes through $\tuple {a \theta, 2 a}$ as was required.

$\blacksquare$

1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Example $3$