Ellipses inscribed in parallelograms $y=\pm 1$, $y=m(x\pm1)$.

Question

Lately I saw many questions about ellipses inscribed in quadrilaterals in Math Stack Exchange. By substituting the equations of lines in the general equation $$ax^2+bxy+cy^2=1$$ of an (maybe) ellipse with symmetry center $(0,0)$ and equating the determinants to zero I found:

1. Ellipses inscribed in the paralelogram $y=\pm1, x=\pm1$ are $$cx^2\pm2\sqrt{c(c-1)}xy+cy^2=1$$ where $c\geq1$.
2. Ellipses inscribed in the parallelogram $y=\pm1, y=mx\pm1$ are $$\frac{m^2c^2}{4(c-1)}-mcxy+cy^2=1$$ where $c>1$.

Questions. A) Are my computations correct? Did İ manage to list all ellipses inscribed in the paralelograms in 1 and 2 above? B) My determinant technique gave complicated equations for the paralelograms $y=\pm1$,$y=m(x\pm1).$ What can I try else?

Answer 1

Any parallelogram with centre $O=(0,0)$ can be described by two consecutive points $P$, $Q$, the other points being their reflections $-P$ and $-Q$ about $O$.

To inscribe an ellipse, choose at will a tangency point $$ T_1=P(1-t)+Qt $$ on $PQ$, where $0<t<1$. The tangency point on $P'Q$ will then be $$ T_2=-P(1-t)+Qt. $$ The midpoint $M$ of $T_1T_2$ lies on $OQ$: $$ M=Qt. $$ A general property of conics states that the line joining center $O$ with the intersection point $Q$ of two tangents, intersects the ellipse at a point $R$ such that $OR^2=OM\cdot OQ$, where $M$ is the midpoint of the tangency points. In our case we can find then a third point on the ellipse: $$ R=Q\sqrt{t}. $$ Plugging then the coordinates of $T_1$, $T_2$ and $R$ into the general equation of an ellipse with center $O$, we get the final result: $$ {(xy_P-yx_P)^2(1-t)+(xy_Q-yx_Q)^2 t \over (x_P y_Q-y_Px_Q)^2(1-t)t} =1. $$ Note that $(x_P y_Q-y_Px_Q)$ is twice the signed area of triangle $OPQ$, hence the above equation can be stated as an equality between squared areas. For instance, in the case $t=1/2$ we can restate it as:

if $A$ is any point on the ellipse, the sum of the squared areas of triangles $AOP$ and $AOQ$ is half the squared area of triangle $OPQ$.

Answer 2

Start by transforming the parallelogram that you have (which is centered at the origin) into a square centered at the origin of side length $2$.

The transformation matrix $L$ applied to the boundary of the parallelogram transform it into a square.

Let $e_1 = (1, 0)$ and $e_2= (0, 1)$

The intersection of $y = -1 $ and $ y = m (x+1)$ is the point $P=( -1 - \dfrac{1}{m} , -1 ) $ and the intersection of $ y = -1 $ and $ y = m (x - 1) $ is the point $Q = (1 - \dfrac{1}{m} , -1 ) $. From symmetry, the reflection of these points about the origin gives the other two vertices $P', Q'$ of the parallelogram.

Let $v_1 = Q - P = (1 - \dfrac{1}{m} , -1) - ( -1 - \dfrac{1}{m} , -1 ) = (2, 0) $

and let $v_2 = Q' - P = (\dfrac{1}{m} - 1, 1 ) - (-1 - \dfrac{1}{m} , -1 ) = (\dfrac{2}{m} , 2 ) $

Now our transformation matrix is such that

$ L v_1 = 2 e_1 $

and

$ L v_2 = 2 e_ 2 $

Therefore,

$ L = 2 [v_1, v_2]^{-1} = 2 \begin{bmatrix} 2 && \dfrac{2}{m} \\ 0 && 2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 && - \dfrac{1}{m} \\ 0 && 1 \end{bmatrix}$

Now pick a point on $PQ$ (in the parallelogram), for example point

$ T = (1 - t) P + t Q = (1-t) ( -1 - \dfrac{1}{m} , -1) + t (1 - \dfrac{1}{m} , -1 ) = ( 2 t - 1 - \dfrac{1}{m} , -1 ) $

The corresponding point in the square is

$T' = L T = (2 t - 1 , -1 ) $

and let this be your tangency point.

The equation of an ellipse in the square that is centered at the origin is

$ r^T A r = 1 $

The gradient vector (which is normal to the ellipse) is

$ N = 2 A r $

At point $T'$ it is pointing in the direction $(- e_2)$. This means that

$ A T' = - k_2 e_2 $

for some $k_2 \gt 0 $

Then

$ T' = - k_2 B e_2 $

where $B = A^{-1} $

Substituting $T'$ into the equation of the ellipse gives us

$ k_2 = \dfrac{1}{\sqrt{ e_2^T B e_2 }} = \dfrac{1}{\sqrt{B_{22}}} $

Now,

$ e_2^T T' = -1 = - \sqrt{ B_{22}} $

Therefore, $B_{22} = 1 $

Also,

$ e_1^T T' = - \dfrac{ e_1^T B e_2 }{\sqrt{B_{22}}} = - B_{12}$

Hence, $B_{12} = 1 - 2 t $

And for similar reasons, $B_{11} = 1 $

So that

$ B = \begin{bmatrix} 1 && 1 - 2 t \\ 1 -2 t && 1 \end{bmatrix} $

So that

$ A = B^{-1} = \dfrac{1}{4 t - 4 t^2 } \begin{bmatrix} 1 && 2 t -1 \\ 2t -1 && 1 \end{bmatrix} $

To get the equation of the original ellipse, we have to make the substitution,

$ r = L r' $

where $r'$ here means the original coordinates. This will give

$ r'^T L^T A L r' = 1 $

Now the matrix $L^T A L$ is given by

$ L^T A L = \dfrac{1}{ 4 t - 4 t^2} \begin{bmatrix} 1 && -\dfrac{1}{m} + 2 t - 1\\ -\dfrac{1}{m} + 2 t - 1 && \dfrac{1}{m^2} - \dfrac{2}{m}(2 t - 1) + 1 \end{bmatrix} $

This completes the specification of the ellipse.

As a numerical example, suppose $m = 2$, $t = 0.25$, then

$ L^T A L = \dfrac{4}{3} \begin{bmatrix} 1 && - 1 \\ -1 && \dfrac{7}{4} \end{bmatrix} $

This is plotted in this Geogebra Worksheet