# Find the angle of rotation and minor axis length of ellipse from major axis length, center, and two points?

I'd like to describe the ellipse centered at the origin with a fixed major axis length of $$2a$$ that passes through two points $$(u_x, u_y)$$ and $$(v_x, v_y)$$ (both within $$a$$ of the origin). In particular, I need to compute the angle of rotation $$\phi$$ of this ellipse, and the length $$2b$$ of its minor axis.

I tried the straightforward approach of just setting up a couple of equations to solve for the coordinates of the foci $$(f_x, f_y)$$ and $$(-f_x, -f_y)$$:

$$\sqrt{(u_x - f_x)^2 + (u_y - f_y)^2} + \sqrt{(u_x + f_x)^2 + (u_y + f_y)^2} = 2a = \sqrt{(v_x - f_x)^2 + (v_y - f_y)^2} + \sqrt{(v_x + f_x)^2 + (v_y + f_y)^2}$$

If I were able to use these equations to determine $$(f_x, f_y)$$ then presumably $$\phi = \arctan(f_y / f_x)$$ and $$b = \sqrt{a^2 - f_x^2 - f_y^2}$$...but solving those equations for $$(f_x, f_y)$$ is beyond my algebraic skills. (They also appear to flummox the SymPy solver.)

• Interesting problem. From playing around with it a bit in GeoGebra, it looks like there are generally two such ellipses. – amd May 3 at 0:41
• I doubt that there’s going to be any “nice” algebraic solution to this, and certainly not a nice general formula. Instead of trying to compute the foci, I would look at the family of conics parameterized by $\theta$ centered at the origin that pass through the two points and $a(\cos\theta,\sin\theta)$. The equations for these are easily constructed with a determinant. De la Hire’s construction of an ellipse seems promising, too: you can construct two inner circles and try to find the angle at which they coincide. – amd May 3 at 0:45

Let $$y=mx$$ be the equation of the major axis of the ellipse, and $$b$$ its semi-minor axis (which lies on line $$y=-x/m$$). Distances $$u_a$$ and $$u_b$$ from point $$u$$ to the lines of the axes are given by: $$u_a^2={(u_y-mu_x)^2\over1+m^2},\quad u_b^2={(mu_y+u_x)^2\over1+m^2},$$ and analogous expressions can be written for point $$v$$. If $$u$$, $$v$$ belong to the ellipse, then: $${u_b^2\over a^2}+{u_a^2\over b^2}=1 \quad\text{and}\quad {v_b^2\over a^2}+{v_a^2\over b^2}=1.$$ This is a system of equations for the unknowns $$m$$ and $$b$$: I solved it with Mathematica and, apart from the trivial solutions $$(b=a,\ m=\pm i)$$, it has two real solutions: $${1\over b^2}= \frac{ u_x^2+u_y^2+v_x^2+v_y^2 -a\left(2u_x^2v_x^2+2u_y^2v_y^2+2 u_x u_y v_x v_y+u_y^2 v_x^2+u_x^2 v_y^2\right) \pm 2 (u_xv_x+u_yv_y)\sqrt{\Delta}} {(u_y v_x-u_x v_y)^2}$$ $$m= \frac{ -u_x u_y \left(a\left(v_x^2+v_y^2\right)-1\right) +v_x v_y\left(a\left(u_x^2+u_y^2\right)-1\right) \mp(u_y v_x-u_x v_y)\sqrt{\Delta}} {u_x^2 \left(1-a v_y^2\right)+v_x^2 \left(a u_y^2-1\right)},$$ where: $$\Delta={\left(1-a \left(u_x^2+u_y^2\right)\right) \left(1-a\left(v_x^2+v_y^2\right)\right)}$$.