Equation of Nephroid

Theorem
Let $H$ be the nephroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the outside of a deferent $C_2$ of radius $a = 2 b$.

Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.

Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.

The point $P = \tuple {x, y}$ is described by the parametric equation:
 * $\begin{cases}

x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$ where $\theta$ is the angle between the $x$-axis and the line joining the origin to the center of $C_1$.

Proof
By definition, a nephroid is an epicycloid with $2$ cusps.


 * Nephroid.png

By Equation of Epicycloid, the equation of $H$ is given by:
 * $\begin{cases}

x & = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta} \\ y & = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta} \end{cases}$

When $a = 2 b$ the equation of $H$ is now given by:
 * $\begin{cases}

x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$