Center of a given ellipse such that it is inscribed in another ellipse

Question

You're given a big ellipse whose axes are parallel to the coordinate axes, with horizontal semi-major axis $a$ and vertical semi-minor axis $b$. And you're also given an ellipse with semi-major axis $c$ and semi-minor axis $d$, where the major axis makes an angle $\theta$ with the positive $x$ axis.

Question: Find the center of the smaller ellipse (of semi-axes $c$ and $d$), such that it is inscribed in the bigger ellipse.

My attempt:

The equation of the big ellipse is

$ r^T Q_1 r = 1 $

where $r = [x, y]^T $ and $ Q_1 = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $

And the equation of the smaller ellipse is

$ (r - C)^T Q_2 (r - C) = 1 $

where $C$ is the center to be determined, and

$ Q_2 = \begin{bmatrix} \dfrac{1}{a^2} \cos^2 \theta + \dfrac{1}{b^2} \sin^2 \theta && \sin \theta \cos \theta \left( \dfrac{1}{a^2} - \dfrac{1}{b^2} \right) \\ \sin \theta \cos \theta \left( \dfrac{1}{a^2} - \dfrac{1}{b^2} \right) && \dfrac{1}{a^2} \sin^2 \theta + \dfrac{1}{b^2} \cos^2 \theta \end{bmatrix} $

Note that $Q_1$ and $Q_2$ are known.

Let $r_1$ be the first tangency point, then

$ r_1 Q_1 r_1 = 1 $

$ (r_1 - C)^T Q_2 (r_1 - C) = 1 $

$ Q_1 r_1 = K_1 \ Q_2 (r_1 - C) $

And let $r_2$ be the second tangency point, then

$ r_2 Q_1 r_2 = 1 $

$ (r_2 - C)^T Q_2 (r_2 - C) = 1 $

$ Q_1 r_2 = K_2 \ Q_2 (r_2 - C) $

These are all the equations, and they are $8$ equations in $8$ unknowns.

These equations can be solved numerically.

And there will be two possible solutions.

Edit:

The above equations will result in infinite solutions, this can be seen by letting $r_2 = r_1$, and eliminating $K_1$, then the unknowns are $r_1$ and $C$ and there are only three equations.

As a way out of this, take $r_1$ on the first ellipse, and compute $C$ from

$ (r_1 - C)^T Q_2 (r_1 - C) = 1 $

$ Q_1 r_1 = K_1 \ Q_2 (r_1 - C) $

Having obtained $C$, now use

$ (r_2 - C)^T Q_2 (r_2 - C) = 1 $

$ Q_1 r_2 = K_2 \ Q_2 (r_2 - C) $

to obtain $r_2$, and then use

$ r_2 Q_1 r_2 = 1 $

as a measure of error in $r_2$.

Answer 1

1): Following figure shows that construction is possible only if the line $y=mx+n$ is perpendicular at big ellipse at one point: a and c represent this condition and b represents other case.

2): As can be seen in following figure the center of small ellipse is where the perpendicular bisector of BC intersects line $y=mx+n$ wich is point A.

You may use this method to find A.We have:

$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1\space\space\space \space\space\space\space\space\space(1)$

$y'=-\frac ba\cdot\frac x{\sqrt{a^2-x^2}}$

$m=-\frac1{y'}$

So we must have:

$m=tan\theta=\frac ab\cdot\frac{\sqrt{a^2-x^2}}x\space\space\space\space\space\space\space\space\space(2)$

(1) and (2) give the coordinates of B. You have the coordinates of C, so you can find the equation of the perpendicular bisector of BC, this equation together with equation $y=mx+n$ gives you the coordinates of A. Note that $y=mx +n$ passes through point B, knowing $m$ and the coordinates of B you can find $n$.The wanted small ellipse is in fact the compressed form of dashed circle.The software I draw the figure draws ellipses based on center and one end point of the major axis, but by Geogebra you need to have focii and a point of the ellipse.

Answer 2

I've implemented the idea suggested by @MathFont in the comments above, to obtain the tangent ellipses (There is two of them).

Step $1$: is to scale up in the $y$ direction by a scale factor $s = \dfrac{a_1}{b_1} $. This transforms the big ellipse into a circle of radius $a_1$. The transformed $Q_2$ is given by $ Q'_2= S^{-T} Q_2 S^{-1} $ where

$ S = \begin{bmatrix} 1 && 0 \\ 0 && \dfrac{a_1}{b_1} \end{bmatrix} $

Step $2$: From $Q'_2$ obtained in the first step, find the semi-major and semi-minor axes of the stretched second ellipse, also find the angle of inclination $\phi$ of the major axis of the transformed second ellipse with respect to the positive $x$ axis.

Step $3$: Place a congruent ellipse having the semi-axes found in Step $2$ such that its major axis is horizontal, centered at the $y$ axis. We want to find the location of its center of the $y$ axis such that it becomes tangent to the big circle which is of radius $a_1$. Let the $Q$ corresponding to this axis aligned version be $Q_3$.

Step $4$: Let $C = [0, c]$ be the center of the this axis-aligned ellipse, and let $r_1$ be the tangency point with the big circle, then

$ r_1^T r_1 = a_1^2 $

$ (r_1 - C)^T Q_3 (r_1 - C) = 1 $

$ r_1^T E Q_2 (r_1 - C) = 0 $

where $E = e_1 e_2^T - e_2 e_1^T $ and $e_1 = [1, 0]^T, e_2 = [0, 1]^T $.

These $3$ equations are quadratic in $r_{1x}, r_{1y}, c $. They can be solved using MATHEMATICA script (or by some other means). We have to exclude the solutions where $r_{1x} = 0$. And also, take only the solutions where $r_{1x} \gt 0 $ (because of mirroring). This will give us only two solutions.

Step $5$: Having obtained the center $ C = [0, c] $, rotate this center by angle $\phi$ to obtain the center of the stretched second ellipse.

Step $6$: The center we want is obtained by applying $S^{-1}$ to the rotated vector $C$ as obtained in Step $5$.

As an example I took the big ellipse to have semi-major axis of $a_1 = 5$ and semi-minor axis of $b_1 = 3$. And the small ellipse to have semi-major axis of $a_2 = 3$ and a semi-minor axis of $b_2 = 1$, and an angle of inclination of the semi-major axis with respect to the positive $x$ axis of $\theta = +45^\circ$.

Applying the above steps, I obtained the following solutions: