Length of Tangent to Astroid between Axes equals Radius of Stator

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Theorem

Let $C_1$ be a circle of radius $b$ roll without slipping around the inside of a circle $C_2$ of radius $a = 4 b$.

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 $H$ be the astroid formed by the locus of $P$.


The segment of the tangent to $H$ between the $x$-axis and the $y$-axis is constant and equals the radius of the stator $C_2$.


Proof

AstroidTangent.png


From Equation of Astroid, $H$ can be expressed as:

$\begin{cases}

x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$

Thus the slope of the tangent to $H$ at $\tuple {x, y}$ is:

\(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {3 a \sin^2 \theta \cos \theta \rd \theta} {-3 a \cos^2 \theta \sin \theta \rd \theta}\)
\(\ds \) \(=\) \(\ds -\tan \theta\)

Thus the equation of the tangent to $H$ is given by:

$y - a \sin^3 \theta = -\tan \theta \paren {x - a \cos^3 \theta}$


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The $x$-intercept is found by setting $y = 0$ and solving for $x$:

\(\ds x\) \(=\) \(\ds a \cos^3 \theta + a \sin^2 \theta \cos \theta\)
\(\ds \) \(=\) \(\ds a \cos \theta \paren {\cos^2 \theta + \sin^2 \theta}\)
\(\ds \) \(=\) \(\ds a \cos \theta\) Sum of Squares of Sine and Cosine


Similarly, the $y$-intercept is found by setting $x = 0$ and solving for $y$, which gives:

$y = a \sin \theta$


The length of the part of the tangent to $H$ between the $x$-axis and the $y$-axis is given by:

\(\ds \sqrt {a^2 \cos^2 \theta + a^2 \sin^2 \theta}\) \(=\) \(\ds a \sqrt {\cos^2 \theta + \sin^2 \theta}\)
\(\ds \) \(=\) \(\ds a\) Sum of Squares of Sine and Cosine

which is constant.

$\blacksquare$


Sources

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