In the attached figures, we have two kinds of curves; 1- a big closed curve which is somehow a limaçon. 2- many ellipses which are the same by the formula but different by the center. By putting some rotate and translated elliple, that is $$4\big(\frac{u}{\tau}\big)^2+\frac{13}{16}\big(\frac{v+\frac13}{\tau}\big)^2+\frac{7}{16}\big(\frac{1}{\tau}\big)^2-\frac{\sqrt{3}}{48}\big(\frac{3v+1}{\tau^2}\big)=1,$$ I found the figure 3. Now to continue I have to consider another rotated ellipse. Thats is I have to put the new rotated ellipse on different points belonging to the last closed curve in figure 3, and then find the envelope. By envelope, it meas the curve which is tangent to all of these ellipses. Supposing that the rotated ellipse is $ax^2+bxy+cy^2=1$ whose centers vary on the points on the last closed curve in figure 3, which are shown in next figures, what would be the equation or location of the curve tangent to all these ellipses? I have no idea how to determine such a curve.
Any help is appreciated in advance.
I should say that
1- I can add as many as ellipses I want.
2- I do not have the equation of the big closed curve, the last curve in fugure 1. In fact, I plotted it with the points which I had from it. I think I can have as many as points on it I want.
My effort
1- First of all, I thought I can find such a curve tangent to all the ellipses by considering the outer vertices of them, but it does not work for the top and bottom of the curve.
Let's assume (as an example) the equation of the ellipse is $x^2+2\,x\,y+3\,y^2=1$. We can write that as $(x+y)^2+(\sqrt{2}\,y)^2=1$. So if we make an affine transformation $$u=x+y,\quad v=\sqrt{2}\,y\tag1,$$ the curve, given by some parametric equations $$x=x(t),\quad y=y(t)\tag2$$ in the $x,y$-plane, becomes $$u=u(t),\quad v=v(t)\tag3$$ in the $u,v$-plane. But in the latter, your ellipse became a circle of radius $1$ ($u^2+v^2=1$), and you can find the envelope simply by computing the normal vector (in the usual, Euklidean geometry of the $u,v$-plane, i.e. with components $\displaystyle\left(\frac{-v'(t)}{\sqrt{u'^2(t)+v'^2(t)}},\frac{u'(t)}{\sqrt{u'^2(t)+v'^2(t)}}\right)$) and going a distance $1$ in the outer or inner direction. Of course, you'll get an outer and an inner envelope. Even if you have your curve only as a sufficiently dense list of points $(x_i,y_i)$, they'll become a list of points $(u_i,v_i)$, and you can approximate the normal vector, there, just using $u_{i+1}-u_i$ instead of $u'(t)$, and $v_{i+1}-v_i$ instead of $v'(t)$. And after finding the envelope(s) in the $u,v$-plane, you transform back via $$x=u-\frac1{\sqrt{2}}\,v,\quad y=\frac1{\sqrt{2}}\,v\tag4.$$