Let’s say I have an ellipse with horizontal axis $a$ and vertical axis $b$, centered at $(0,0)$.
I want to compute $a’$ and $b’$ of a smaller ellipse centered at $(0,0)$, with the axes rotated by some angle $t$, tangent to the bigger ellipse and $\frac{a’}{b’}=\frac{a}{b}$.
Given ellipse of axes $a$ and $b$, find axes of tangential and concentric ellipse at angle $t$
conic sectionsgeometry
Related Solutions
What I'd do
I read up to $\phi=\psi-\alpha$ and agree with that. I'm spinning my own thoughts from that point on. For the moment I'd like to think of your tangent direction not as an angle but as a direction vector instead, namely the vector
$$ v_1 = \begin{pmatrix} \cos\phi \\ \sin\phi \end{pmatrix} $$
Now you can take your whole scene and scale all $x$ coordinates by $a$ and all $y$ coordinates by $b$. This will turn your ellipsis into a unit circle, and the tangent direction will become
$$ v_2 = \begin{pmatrix} \frac{\cos\phi}a \\ \frac{\sin\phi}b \end{pmatrix} $$
Now you're looking for the point on the unit circle with this tangent. This is particularly easy, since tangents to the unit circle are perpendicular to radii. Simply take your vector, swap $x$ and $y$ coordinate and also swap one sign. That will result in a perpendicular vector
$$ v_3 = \begin{pmatrix} \frac{\sin\phi}b \\ -\frac{\cos\phi}a \end{pmatrix} $$
Now you have to change the length of that vector to $1$ so you get a point on the unit circle.
$$ v_4 = \frac{v_3}{\lVert v_3\rVert} = \frac{1}{\sqrt{\left(\frac{\sin\phi}b\right)^2 + \left(\frac{\cos\phi}a\right)^2}} \begin{pmatrix} \frac{\sin\phi}b \\ -\frac{\cos\phi}a \end{pmatrix} = \begin{pmatrix} x_4 \\ y_4 \end{pmatrix} $$
Note that at this point, the opposite point $-v_4$ is a second valid solution. Now you can scale your coordinates back by $a$ and $b$ and end up with
$$ v_5 = \begin{pmatrix} a\cdot x_4 \\ b\cdot y_4 \end{pmatrix} $$
Last you'd apply the rotation by $\alpha$ to that (and possibly $-v_5$ as well).
What you did
So now that I've thought about how I'd think about this, I'll have a look at the rest of what you did. It seems that your computations look a lot shorter than mine, so they might be more efficient for practical uses. Nevertheless, my approach might yield some insight as to what the individual steps do, so I'll leave it in place and even refer to it.
I found that I had to move that $−$ from the $y$ part to the $x$ part otherwise my calculation was off by $180°$
If your tangents are unoriented lines, then a change in $180°$ in that argument will give an equally valid result. This is the $v_4$/$-v_4$ ambiguity in my solution. The “velocity vector” you used is oriented, pointing in a given direction, but if you move along your circle in the opposite direction, you'd get opposite velocities at the same points.
So again the question: how does the above solution work, or what is a better more proper rigorous derivation of a solution to the problem?
Your solution looks good. You might want to consider $t$ and $t+180°$, and if you do, then it shouldn't matter where you place your minus sign. If you still have doubts, however, feel free to implement my approach as an alternative and compare the results. They should agree.
I should also note that http://mathworld.wolfram.com/Ellipse.html eqn 60 gives the relation between $\phi$ and $t$ as […]
Their $\phi$ is the “polar angle” of a point on the ellipsis. Look at the line connecting the center of the ellipsis with a given point on the ellipsis. The angle that line makes with the $x$ axis is their $\phi$.
$$\large 4(a^2\sin^2\theta+b^2\cos^2\theta)=w^2$$ Equation of an ellipse is: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Diffrentiate: $$\frac{dy}{dx}=-\frac{b^2}{a^2}.\frac{x}{y}$$ Polar form of ellipse: $$P(\phi)\equiv(a\cos\phi,b\sin\phi)$$ Slope of tangent in polar form: $$m=-\frac ba\cot\phi$$ Equation of tangent: $$\frac xa\cos\theta+\frac yb \sin\theta=1$$ Distance between two parallel lines in form: $$ax+by+c=0\\ax+by+c'=0$$ is: $$d=\frac{|c-c'|}{\sqrt{a^2+b^2}}$$
Firstly those two lines are tangent at two points, and they are parallel too. So in the parametric form the slope of tangent at these two points – let them be called $P(\theta_1),Q(\theta_2)$ – must be equal.
- slope $\propto\cot\phi$, where $\phi$ is parametric angle. $$\implies \cot \theta_1=\cot\theta_2\implies \theta_1=\pi+\theta_2\quad\text{since}\quad\theta_1\ne\theta_2$$
Secondly the distance between the two tangents / two parallel lines must be $w$:
- Tangent Lines are $$\frac xa\cos\theta_1+\frac yb\sin\theta-1=0\\\frac xa\cos\theta+\frac yb\sin\theta+1=0$$ (after putting $\theta_2=\pi+\theta_1$). So taking distance between them: $$\frac{|1-(-1)|}{\sqrt{\frac{\cos^2\theta_1}{a^2}+\frac{\sin^2\theta_1}{b^2}}}=w$$
Thirdly slope of tangents at these points is $\theta$ so: $$-\frac{b^2}{a^2}\cot\theta_1=\tan\theta$$ Try the rest yourself.
Best Answer
Let the ratio $r= \frac{a’}a=\frac{b’}b$. Then, the inscribed ellipse with the tilt angle $t$ is
$$\frac{(x\cos t+y\sin t)^2}{r^2a^2}+ \frac{(-x\sin t+y\cos t)^2}{r^2b^2}=1\tag 1 $$
Also, the inscribed ellipse can be expressed as
$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}-k\left( \frac{x\sin \theta}{a}-\frac{y\cos \theta}{b}\right)^2=1\tag 2$$
as pointed out by @Ng Chung Tak. Match the terms of $x^2$, $y^2$ and $xy$ between (1) and (2) to establish the system of equations below
\begin{align}k\sin 2\theta &=\frac{b^2-a^2}{r^2ab}\sin 2t\\ 1-k\sin^2 \theta &= \frac1{r^2b^2}(b^2\cos^2 t+a^2\sin^2t)\\ 1-k\cos^2 \theta &= \frac1{r^2a^2}(a^2\cos^2 t+b^2\sin^2t)\\ \end{align}
Plug the 1st equation into the others to eliminate $k$
$$r^2- \frac{b^2\cos^2 t+a^2\sin^2t}{b^2}= \frac{b^2-a^2}{2ab}\sin 2t\tan\theta $$
$$r^2- \frac{a^2\cos^2 t+b^2\sin^2t}{a^2}= \frac{b^2-a^2}{2ab}\sin 2t\cot\theta $$
Then, multiply the two equations to eliminate $\theta$. After simplification $$r^4 -( 2+\Delta^2)r^2+1=0$$
where $\Delta = \left( \frac a b - \frac b a\right)\sin t$. Solve to obtain the axis ratio
$$r= \sqrt{1+\frac{\Delta^2}4}-\frac{\Delta}2$$
Thus, the axes of the smaller ellipse are
$$a’ = ra, \>\>\>\>\>b’= rb$$
As expected, $r=1$ in the special case of $t=0$, and $r=\frac b a$ of $t=\frac\pi2$.