What is the relation among sides of triangle in oval (ellipse) if longest side of triangle is the longest diameter of oval, the rest point is located on the side of oval, you can vary it to change the triangle and except the longest side, set one of side of triangle be variable t how to the find another side in terms of t? if needed, you can add other variables beside longest side of triangle is the longest diameter of oval (ellipse), there is another case is that longest side of triangle is the length of between two focus of oval (ellipse), if possible, use least number of variables to find another side, great one is another side's equation is in terms of only t. if impossible, then please show other variables in a diagram Remark: this remark question is not the required one, if can not show in one variables, will it mean that the two sides that can have different equations if in terms of two or more variables? what are they?

!Ellipse

Let $a > b$ be the semi-axes of the ellipse and $c = \sqrt{a^2 - b^2}$ be half of the distance between the two foci. With a suitable choose of origin and coordinate axes, one can make the ellipse satisfies:` $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad\iff\quad \- \frac{c^2}{a^2b^2} x^2 + \frac{1}{b^2}(x^2 + y^2) = 1\tag{*1} $$

Let $s$ be the length of the other side of your triangle, one have:

$$\begin{cases} t^2 = (x-a)^2 + y^2\\\ s^2 = (x + a)^2 + y^2 \end{cases} \implies \begin{cases} s^2 - t^2 = 4ax\\\ s^2 + t^2 = 2(x^2 + y^2 + a^2) \end{cases} $$

Substitute this back into $(*1)$, one obtain a quadratic equation in $s^2$:

$$-\frac{c^2}{a^2b^2}\left(\frac{s^2 - t^2}{4a}\right)^2 + \frac{1}{b^2}\left(\frac{s^2 + t^2}{2} - a^2\right) = 1$$

Solving this give us: $$s = \sqrt{\frac{4 a^2 \sqrt{c^2 t^2+b^4}+ c^2 t^2 +4a^4}{c^2}}$$