Question

How to determine an ellipse passing by the four points `p_1 = {5, 10},p_2 = {-2, -10};p_3 = {-5, -4};p_4 = {5, -5};`?

Answer 1

See below.

Explanation:

A general conic has the form

`C=a x^2+b y^2+c x y + d x + e y + f=0` it can be read also as

`C = ((x),(y))((a,c/2),(c/2,b))(x,y)+(d,e)((x),(y))+f`

We know also that the kind of conic is dictated by the matrix

`M=((a,c/2),(c/2,b))`

characteristic polynomial.

`p_M(lambda)=lambda^2-2(a+b)lambda+c^2-4ab`

The ellipses are those conics for which the characteristic polynomial roots are real with same sign. We consider that the circle is a particular case of ellipse.

Keeping that in mind the condition is

`(a - b)^2 + c^2< (a+b)^2` or simplifying

`ab > (c/2)^2`

Now, substituting `p_1,p_2,p_3,p_4` into `C` we have

`{(25 a + 100 b + 50 c + 5 d + 10 e + f=0),(4 a + 100 b + 20 c - 2 d - 10 e + f=0),(25 a + 16 b + 20 c - 5 d - 4 e + f=0),(25 a + 25 b - 25 c + 5 d - 5 e + f=0),(a b = (c/2)^2 +delta^2):}`

The last equation is the "ellipse" condition with `delta^2>0` assuring the inequality.

Here we have `5` equations and `7` incognitas.

Solving for `a,c,d,e,f` we obtain

`((a = 1/35 (463 b - 4 sqrt[35] sqrt[323 b^2 - 35 delta^2])),(c = 2/35 (70 b - sqrt[35] sqrt[323 b^2 - 35delta^2])),(d = 1/35 (511 b - 8 sqrt[35] sqrt[323 b^2 - 35 delta^2])),(e = 1/7 (-175 b + 2 sqrt[35] sqrt[323 b^2 - 35 delta^2])),(f = 4/7 (-794 b + 7 sqrt[35] sqrt[323 b^2 - 35 delta^2])))`

so we have infinite solutions depending on

`323 b^2 - 35delta^2 ge 0`

so choosing `b=b_0` we have

`0 le delta^2 le 323/35 b_0^2`

so

`C = C(delta)`

Attached a plot showing some such ellipses.