# 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};?

Feb 14, 2017

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

$C = \left(\begin{matrix}x \\ y\end{matrix}\right) \left(\begin{matrix}a & \frac{c}{2} \\ \frac{c}{2} & b\end{matrix}\right) \left(x , y\right) + \left(d , e\right) \left(\begin{matrix}x \\ y\end{matrix}\right) + f$

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

$M = \left(\begin{matrix}a & \frac{c}{2} \\ \frac{c}{2} & b\end{matrix}\right)$

characteristic polynomial.

${p}_{M} \left(\lambda\right) = {\lambda}^{2} - 2 \left(a + b\right) \lambda + {c}^{2} - 4 a b$

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

${\left(a - b\right)}^{2} + {c}^{2} < {\left(a + b\right)}^{2}$ or simplifying

$a b > {\left(\frac{c}{2}\right)}^{2}$

Now, substituting ${p}_{1} , {p}_{2} , {p}_{3} , {p}_{4}$ into $C$ we have

$\left\{\begin{matrix}25 a + 100 b + 50 c + 5 d + 10 e + f = 0 \\ 4 a + 100 b + 20 c - 2 d - 10 e + f = 0 \\ 25 a + 16 b + 20 c - 5 d - 4 e + f = 0 \\ 25 a + 25 b - 25 c + 5 d - 5 e + f = 0 \\ a b = {\left(\frac{c}{2}\right)}^{2} + {\delta}^{2}\end{matrix}\right.$

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

$\left(\begin{matrix}a = \frac{1}{35} \left(463 b - 4 \sqrt{35} \sqrt{323 {b}^{2} - 35 {\delta}^{2}}\right) \\ c = \frac{2}{35} \left(70 b - \sqrt{35} \sqrt{323 {b}^{2} - 35 {\delta}^{2}}\right) \\ d = \frac{1}{35} \left(511 b - 8 \sqrt{35} \sqrt{323 {b}^{2} - 35 {\delta}^{2}}\right) \\ e = \frac{1}{7} \left(- 175 b + 2 \sqrt{35} \sqrt{323 {b}^{2} - 35 {\delta}^{2}}\right) \\ f = \frac{4}{7} \left(- 794 b + 7 \sqrt{35} \sqrt{323 {b}^{2} - 35 {\delta}^{2}}\right)\end{matrix}\right)$

so we have infinite solutions depending on

$323 {b}^{2} - 35 {\delta}^{2} \ge 0$

so choosing $b = {b}_{0}$ we have

$0 \le {\delta}^{2} \le \frac{323}{35} {b}_{0}^{2}$

so

$C = C \left(\delta\right)$

Attached a plot showing some such ellipses.