Consider this

Let $p$ be the $x$ -coordinate of the vertex that slides along the $x$ -axis and $q$ be the $y$ -coordinate of the vertex that slides along the $y$ -axis.

Since the two green triangles pictured above are congruent by AAS congruence, the third vertex can be defined by $x = p + q$ and $y = p$ . Also, since $p$ and $q$ are legs of a right triangle with a unit hypotenuse, $q = \sqrt{1 - p^2}$ . Combining these equations gives $x = y + \sqrt{1 - y^2}$ which in standard form is $x^2 - 2xy + 2y^2 - 1 = 0$ .

The angle $\theta$ that the major axis makes with the positive $x$ -axis is the same as the angle of rotation which is defined by $\cot 2\theta = \frac{A - C}{B}$ , and since $A = 1$ , $B = -2$ , and $C = 2$ , this simplifies to $\tan 2\theta = 2$ . Using the double angle formula for tangent, $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = 2$ , which solves to $\tan \theta = \frac{-1 + \sqrt{5}}{2} = \boxed{\phi - 1}$ .