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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.

Figure 3

enter image description here

Rushabh Mehta
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Majid
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  • Do you have an equation for the "closed curve which is somehow a limaçon"? – qfwfq Mar 08 '21 at 11:02
  • Your attempt to solve it? Any ideas? – Cesareo Mar 08 '21 at 12:00
  • @qfwfq: Unfortunately, I do not have its equation, but as many as points on it I want. In fact, I obtained its curve starting from a limaçon, whose equation is at hand, adding two vectors on each point of it, finding the outer vector, and plotting a curve which pass through the head of those vectors. In fact, it is a curve which predicts the trajectory of wildfire spreading. As I thought it might make the things more complicated, I did not give these details in the question. But, I add a sentence to question to say what I have about the big closed curve! Thank you so much for your attention! – Majid Mar 08 '21 at 12:52
  • @Cesareo: I do not have any right idea to handle it, but I add some incomplete efforts which I have tried so far. Thanks for your attention! – Majid Mar 08 '21 at 12:54
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    @Majid: You're looking for the "envelope" of the ellipses. Not knowing the eqn of the big curve makes this trickier, but presumably the corresponding differential equations could be solved numerically from your data points. ... "I obtained its curve starting from a limaçon, whose equation is at hand, adding two vectors on each point of it, finding the outer vector, and plotting a curve which pass through the head of those vectors." Please provide exact details of this in the question; it may be possible to derive the big curve's eqn. – Blue Mar 08 '21 at 13:47
  • Your words "adding two vectors on each point of it, finding the outer vector" sound as if it were possible to do that analytically, in principle, to find an equation (a parametric one, most likely) for the curve. There's a reason for that: the "canonical" way of finding an envelope for a (parametric) family of curves includes a partial derivative with respect to that parameter, and that's not possible if you have only discrete points. Well, I guess that @Blue is saying the same, and it may be possible to replace that partial derivative by the difference of neighboring points on the curve. – qfwfq Mar 08 '21 at 20:04
  • Thanks erevyone! I edit the expression of my question and put more informaion. But exactly, as @ Blue and @qfwfq mentioned I am trying to find the envelope. – Majid Mar 08 '21 at 20:28
  • Good job! But I think I found a simpler way to get your envelope, as long as your curve is sufficiently regular. I'll add an answer, now. – qfwfq Mar 09 '21 at 06:59

1 Answers1

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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 $$x=u-\frac1{\sqrt{2}}\,v,\quad y=\frac1{\sqrt{2}}\,v\tag4.$$

qfwfq
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  • by normal vector you mean the vector which is orthogonal to the last closed curve in Fig 3? If so, I do not know how to figure out the orthogonal vectors as I do not have the equation of this curve. Also the metric here is some kind of Euclidean, not necsesarily the standard one. That is it is not so easy to find the orthogonal vectors. – Majid Mar 09 '21 at 18:21
  • @Majid No, I've edited my post to make it (hopefully) clearer. – qfwfq Mar 09 '21 at 19:29