Are subnormal and polar subnormal equal?

I have been studying Jacobi Elliptic functions and I am trying to understand the proof that the curve which ellipse describes is of the form $dn(x)$.I found this proof:

Where there is written $MG = - \frac{dr}{d\theta} = - y \frac{dy}{dx}$ this are supposed to be subnormal and polar subnormal. Now in polar coordinates $\frac{dy}{dx} = \frac{r’\sin{\theta} + r\cos{\theta}}{r’\cos{\theta} - r\sin{\theta}}$ however I could not figure out why is $\frac{dr}{d\theta} = y \frac{dy}{dx}$ true since definition of subnormal and polar subnormal are different so it must be true in this particular set up. Can anyone clarify?

Solution 1:

You want to show that

$$MG=-r' \tag{1}$$

(with notation $r':=\frac{dr}{d\theta}$).

It is a consequence of the (rather classical) formula

$$\tan \psi = \dfrac{r}{r'}\tag{2}$$

where $\psi$ is the (oriented) angle between the radius-vector $OP$ and the tangent at point $P$.

Formula (2) is for example mentionned in the answer by Nosrati to this (interesting) question

Now, with the notations you find below in your "augmented" figure :

$$\tan \psi = \tan(\alpha + \pi/2) = - \frac{1}{\tan \alpha} = -\frac{r}{MG}\tag{3}$$

(the last inequality being established in triangle $PMG$).

Setting equality between (2) and (3) gives (1).

Remark: The subnormal $MG$ of the curve is indeed $-yy'$ (a classical formula).

Different questions [in fact answered by having access to the (Google) book of Sir Greenhill named "Applications of elliptic functions"]:

• You say "the curve that the ellipse describes". Could it be said a stronger property, i.e., "the curve on which the ellipse is rolling without slipping" (in such a way that its center moves on a horizontal line) ? [Answer $\to$ yes]

• What is the relationship between parameter $m$ of the elliptic curve $y=\operatorname{dn}(x,m)$ and the values of $a,b$ ? [Answer $\to$ in fact it is $y = a \operatorname{dn}(x/b,e)$ with $m = e $ (the ellipse's eccentricity)]

Edit: About the point you are underlining, I see a classical cause of misunderstanding: one should never use the same letter as the integration variable AND an integration bound like here:

$$x=\int_y^{a} \dfrac{ab dy}{\sqrt{(a^2-y^2)(y^2-b^2)}}$$

but INSTEAD write this under the form:

$$x=\int_{Y=y}^{Y=a} \dfrac{ab dY}{\sqrt{(a^2-Y^2)(Y^2-b^2)}}dY$$ where the integration variable (capital) $Y$ is distinguishable from the bound $y$, making the limit case (brough by extremal value $y=b$) much more natural:

$$\int_{Y=b}^{Y=a} \dfrac{ab dY}{\sqrt{(a^2-Y^2)(Y^2-b^2)}}dY$$