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EllEvoluteProp

PM is a segment of ellipse normal upto intersecting point $M$ on y- axis and PC is the radius of curvature.

EDIT1:

Show that (segment lengths of) PC is proportional to the cube of PM, the constant of proportionality being $ (b^2/a^4)$ at all points $P$. Two locations are verified as under:

At $a$:

$$ \frac{PC}{PM^3}=\frac{b^2/a}{a^3}=\frac{b^2}{a^4} $$

At $b:$

$$ \frac{PC}{PM^3}=\frac{a^2/b}{(a^2/b)^3}= \frac{b^2}{a^4} $$

Narasimham
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  • This can be solved with a straightforward application of the radius of curvature formula applied to, say, the standard parameterization $(a\cos\theta, b\sin\theta)$. What aspect of that approach is giving you difficulty? – Blue Nov 19 '19 at 09:32
  • The fact that $M$ is on the major axis doesn't matter much. (That is, nothing about the problem hinges on the fact that $M$ is collinear with the foci of the ellipse.) A comparable result holds when $M$ is on the minor axis; the roles of the major and minor radii are simply reversed in the target proportion. – Blue Nov 19 '19 at 10:10
  • Yes, apologies. you are right. it was a simple graphic error, now labelled point M on y-axis correctly. All else okay. – Narasimham Nov 19 '19 at 17:34
  • It's not clear what you want. You apparently know how to derive the constants of proportionality. So ... What's your question? (If you're looking for an alternative to your own approach, please document your approach so that readers know what "alternative" might mean.) – Blue Nov 19 '19 at 17:36
  • Thanks. From what I remember of the formula I verified at two points. But it should be checked as a property at all points P. To me it is not that straight-forward. – Narasimham Nov 19 '19 at 17:45
  • Since curvature is involved added ODE tag. – Narasimham Nov 19 '19 at 17:51

1 Answers1

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Take the standard parameterization of the ellipse:

$$P := (a\cos t, b\sin t) \tag{1}$$ where $a$ and $b$ are the "horizontal" and "vertical" radii, not necessarily "major" and "minor". (The major/minor distinction is immaterial.) To avoid sign complications, we'll consider the first-quadrant arc of the ellipse, where $0\leq t\leq \pi/2$. With this, we are assured that an "inward-pointing" normal at $P$ is given by $$n := (-b \cos t, -a \sin t) \tag{2}$$ (which is obtained from exchanging the components of the tangent vector $P'(t)$, and changing signs to ensure the proper orientation). A point $K$ at distance $k$ from $P$ along the normal line has the form $$K := P + \frac{k}{|n|} n = \left(\; \left(a-\frac{bk}{|n|}\right) \cos t,\;\left(b-\frac{ak}{|n|}\right)\sin t\;\right) \tag{3}$$ In particular, setting appropriate coordinates to zero, we find that the point $X$ and $Y$ on the $x$- and $y$-axes corresponds to the distances $$|PX| =\frac{b}{a}|n| \qquad |PY| = \frac{a}{b}|n| \tag{4}$$ Now, the point $Z$ on the evolute has distance from $P$ equal to the radius of curvature of the ellipse at $P$. By the parametric formula, we have

$$|PZ| := \frac{\left(P_x'^2 + P_y'^2\right)^{3/2}}{\left|P_x'' P_y'-P_x'P_y''\right|} = \frac{|n|^3}{a b} \tag{5}$$

(where I'm using $P_x$ and $P_y$ to refer to the coordinates of $P$). Thus,

$$|n|^3 = \frac{a^3}{b^3}|PX|^3 = \frac{b^3}{a^3}|PY|^3 = ab|PZ| \tag{$\star$}$$

and the result follows. $\square$

Blue
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  • So should the proportionality constant be $\dfrac{1}{ab} ?$ – Narasimham Nov 20 '19 at 14:28
  • @Narasimham: The proportionality constants are the same as you gave. (I merely changed some point names.) That is, from $(\star)$ we can write $$\frac{|PZ|}{|PX|^3} = \frac{a^2}{b^4} \qquad\qquad \frac{|PZ|}{|PY|^3} = \frac{b^2}{a^4}$$ – Blue Nov 20 '19 at 14:55