$(A,C,D,E)$ are four fixed points on a fixed outer (red) ellipse and $B$ is a variable point lying between $A$ and $C$. Alternate vertices are joined to form a pentagonal star. (drawn on Geogebra).

A variable green inner ellipse circumscribing pentagon $(abcde) $ is formed by connecting five internal points of intersection. The green ellipse size and orientation, eccentricity etc. geometric properties are functions of position of $B$ on arc $AC$ as shown. If $B$ does not lie between $(A,C)$ then the green ellipse is not formed at all.

How are equations of red and green ellipses related with respect to the moving point $ B ? $ Do they have any common geometrical property? Can one property be obtained from the other?

Thanks in advance.

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  • $\begingroup$ A failed approach: $(a^2-a)(y^2-y)=(b^2-b)(x^2-x)$ is the conic through $(0,0),(1,0),(0,1),(1,1),(a,b)$. If $0<a<1$ and $b>1$ then the five new points are $(\frac12,\frac12),(\frac{a}{b},1),(\frac{b-a+1}{b},1),(\frac{b}{b-a+1},\frac{b}{b-a+1}),(\frac{a}{b+a},\frac{b}{b+a})$ which from the 6 by 6 determinant yields a unwieldy new conic equation in terms of a and b. $\endgroup$ – Jan-Magnus Økland Jul 10 '18 at 9:02
  • $\begingroup$ The third point should be $(\frac{b-a}{b},1)$ and then the equation becomes $-(a-1)b^2x^2+(2a-1)b(b-a)xy+a(b^2-2ab+b+2a^2-2a)y^2-ab(b-2a+1)x-a(2b^2-2ab+a^2-a)y+ab(b-a)$ $\endgroup$ – Jan-Magnus Økland Jul 11 '18 at 9:47
  • $\begingroup$ It appears something is nicely coming up. May be if another symbol like $c$ for $1$ is used a homogeneous equation would show up with symmetry? $\endgroup$ – Narasimham Jul 11 '18 at 17:16
  • $\begingroup$ I don't fully understand your question. Are you starting with a conic $C_1$, or are you starting with five distinguished points defining $C_1$? If it is five points, I can see how pairwise intersections would lead to a second conic, but it feels as though the conics were mostly defined in terms of their points, with little connection to be expected between $C_1$ and $C_2$. If all you have is the conic $C_1$, I fail to see how $C_2$ is defined from that, since picking five different points would lead to a different conic. Can you clarify this? $\endgroup$ – MvG Jul 12 '18 at 23:56
  • $\begingroup$ I am replacing the question to a much narrower context, May be it could be generalized later. Please indicate if it makes sense. $\endgroup$ – Narasimham Jul 13 '18 at 7:39

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