Normal to Cycloid passes through Bottom 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 normal to $C$ at a point $P$ on $C$ passes through the bottom of the generating circle of $C$.
Proof 1
From Normal to Cycloid, the equation for the normal to $C$ at a point $P = \tuple {x, y}$ is given by:
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Proof 2
From Tangent to Cycloid passes through Top of Generating Circle, the tangent to $C$ at a point $P = \tuple {x, y}$ passes through the top of the generating circle.
By definition, the normal to $C$ at $P$ is perpendicular to the tangent to $C$ at $P$.
From Thales' Theorem, the normal, the tangent and the diameter of the generating circle form a right triangle.
Thus the normal to $C$ at $P$ meets the generating circle at the opposite end of the diameter to the tangent.
Hence the result.
$\blacksquare$