Determine offset of a point on the major axis of an ellipse

Question

I am trying to solve the following integral equation in order to show that the point P on the diagram below is a focus.

$$N = \tfrac{1}{R^2} \int_0^{2\pi} q^2 \cos \phi \,{\rm d}\phi$$

The polar equation for the ellipse in terms of $q$ and $\phi$ is:

$$\left(\frac{q}{R}sin\phi\right)^2 + \left(\frac{q}{R}cos\phi - \frac{b}{R}\right)^2sin^2\beta = 1$$

I would like the solution for the constant "b".

I am hoping that if this can be solved that I can then solve the situation for P' in the diagram which is not a focus of the ellipse. In this case b in the ellipse polar equation would be replaced by b'. Ultimately I am trying to solve for b'. If the solution for b' is in terms of b that is great. If necessary, the assumption that $b' = b/A$, where $A <= 1$ is probably okay.

ADDITIONAL INFORMATION: For the case where we are looking at P (which is a focal point of the ellipse), the following is true:

$$N = 2\pi\cot \beta = \frac{1}{R^2} \int_0^{2\pi} q^2 \cos \phi \,{\rm d}\phi$$

The above equation is really a momentum balance in the "x" direction for a 100% elastic collision (no energy losses at impingement) of a jet impinging against a surface or another equal jet. From a paper, which states that the integral equation and polar equation were used to solve for $b$ but provides no details whatsoever, the solution for b is:

$$b = R\cot\beta$$

What I am trying to do is to solve for $b'$ when the problem instead involves P'. For P', this is a case where the momentum equation is for an impingement that is not 100% elastic (i.e. there is a loss of energy at impingement). The derived equation for this case is as follows:

$$N = \frac{2\pi\cot \beta}{A} = \frac{1}{R^2} \int_0^{2\pi} q'^2 \cos \kappa \,{\rm d}\kappa$$

"$A$" is constant for a specific value of $\theta$. For a 100% elastic collision $A$ is equal to 1 and in that case, $b' = b$, but not if $A < 1$.

My thinking is that if a solution for $b$ can be obtained from the integral for the case where P is a focus by directly solving the integral, then it might be possible to extend it to the case for P'. I suspect that the solution for $b'$ for the P' case is:

$$b' = \frac{R\cot\beta}{A}$$

Perhaps there is a better way to achieve what I am looking for rather than trying to directly solve the integral. It can be assumed that $b = R\cot\beta$ is correct and maybe use this to come up with $b'$, without trying to directly solve the integral.

Answer 1

• In a special case of focal origin,

$$r(\theta)=\frac{b^2}{a-c\cos \theta}$$

• Linear eccentricity

$$c=\sqrt{a^2-b^2}$$

• Area swept \begin{align} A(\theta) &= \frac{1}{2} \int_0^\theta r^2 d\theta \tag{$0 \le \theta \le 2\pi$} \\[4pt] &= ab\cot^{-1} \left( \sqrt{\frac{a-c}{a+c}} \cot \frac{\theta}{2} \right)+\frac{b^2c\sin \theta}{2(a-c\cos \theta)} \end{align}

• Avoiding discontinuity, takes

$$\cot^{-1} (\cdot) \equiv \frac{\pi}{2}-\tan^{-1} (\cdot)$$

• Please also see another posts in Physics SE here and here.

Answer 2

For the equation of the ellipse (which correctly matches the sketch) you have

$$q = \frac{b R^2 \cos \phi}{R^2+(b' \sin\phi)^2} + \sqrt{ \frac{R^2 (R^2+b'^2)(R^2-(b^2-b'^2)\sin^2\phi}{(R^2+b'^2\sin^2\phi)^2} } \tag{1}$$

where $$b' = \sqrt{ \left( \tfrac{R}{\sin\beta} \right)^2-R^2} = R \cot \beta$$ is the focal point.

Now the integral

$$ N = \frac{1}{R^2} \int q^2 \cos \phi {\rm d}\phi $$

$$\small N = \frac{1}{R^2} \int_0^{2\pi} \left(\frac{R^2 b \cos \phi}{R^2+(b' \sin\phi)^2}+\sqrt{ \frac{R^2 (R^2+b'^2)\left(R^2 - (b-b')\sin\phi)^2 \right)}{(R^2+(b' \sin\phi)^2)^2} } \right)^2 \cos \phi {\rm d}\phi \tag{2} $$

does not have an analytical solution because it is an elliptical integral.

But you can actually get a Taylor series expansion of the integral in terms of $b$ near the $b'$ value

This is

$$ N \approx \frac{1}{R^2} \left( \frac{ 2 \pi R \sqrt{R^2+b'^2}(b-b')+2 R^2 (b'-b)+b'^3)}{b'^2} \right) \tag{3}$$

If you evaluate the above at $b = b'$ then (3) is

$$ N = \frac{1}{R^2} \left( \frac{2\pi R ( 0 + 0 + b'^3)}{b'^2} \right) = \frac{1}{R^2} (2\pi R b'^2) = \frac{2\pi b'}{R}$$