Parameters defined for the asymptotes of a polar curve $r(\theta)$

Question

Given a polar curve $\rho(\theta)$,then the asymptote of the curve can be found by the two following parameters: $$\alpha=\lim_{\rho \to \infty}φ\;\;\;\;\;,\;\;\;\;p=\lim_{\rho \to \infty}\rho\sin(\alpha-φ)$$

Consider the following picture:

My question is that where do these parameters come from?

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Answer 1

If you can have a form $\varphi=\varphi(\rho)$, then $\alpha=\displaystyle\lim_{\rho \to \infty}\varphi$ (which should be intuitively obvious) is useful. However, it is sometimes difficult to have a form $\varphi=\varphi(\rho)$. In that case, you can find $\alpha$ by solving $\frac{1}{\rho(\varphi)}=0$ for $\varphi$. For example, if we are given $\rho\sin(\varphi)=2\cos(2\varphi)$, then solving $\frac{\sin(\varphi)}{2\cos(2\varphi)}=0$ gives $\alpha=n\pi$ where $n$ is an integer.

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Let $N$ be a point on the line passing through the pole with the angle $\alpha$ (measured from positive $x$ axis) such that $ON\perp MN$.

Then, we get $$\frac{MN}{OM}=\sin\angle{MON}$$ from which we have $$MN=OM\sin\angle{MON}=\rho\sin(\alpha-\varphi)$$

Therefore, we get $$p=\lim_{\rho\to\infty}MN=\lim_{\rho\to\infty}\rho\sin(\alpha-\varphi)$$

Answer 2

It was assumed that that the student is familiar with straight line equation in polar form. This tries to link the polar spiral or curve's asymptotic behaviour to understand it in terms of the already known elementary straight line polar form parameters $(p, \alpha)$ or $(p,\beta)$.

Equations of given polar curve and straight line asymptote of the polar curve where $OM =\rho$ is a part segment of $OL=r$ are both in polar equation form:

$$ M= f(\rho,\varphi); \;L= g(r,\varphi);\; $$

Equation of straight line OL in standard polar form where $p$ is minimum pedal distance to the straight line and $\beta$ is angle measured from positive $x$ axis to the pedal distance vector in anticlockwise direction $ \beta= (\pi/2- \alpha)\; $:

$$ p =x \cos \beta + y \sin \beta $$

Since $\beta $ is measured clockwise here,

$$ p =x \cos \beta - y \sin \beta $$

To express $OL$ from Cartesian form into polar coordinate form with radius vector $OL= r $ and polar coordinate $\varphi:$

$$ p = r\cos \varphi \cos \beta- r \sin \varphi \sin \beta = r \cos (\varphi +\beta) $$

It can be also expressed

$$ p= r \cos ( \varphi +(\pi/2- \alpha)) = r \sin (\varphi + \alpha) $$

$$ p= r \sin \psi $$

This is the Clairaut's minimum distance of tangent to any curve from pole/origin. The straight line is a simple special curve case.

EDIT1:

Basically how do you get the pedal distance $p -\gamma$ form of a straight line? In the following general derivation for another angle $\gamma$ it is shown measured in positive counterclockwise direction.