Confocal Conics are Self-Orthogonal

Theorem
The confocal conics defined by:


 * $\quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$

forms a family of orthogonal trajectories which is self-orthogonal.


 * ConfocalConics.png

Proof
Consider:
 * $(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$

From Equation of Confocal Ellipses: Formulation 2:
 * $(1)$ defines an ellipse when $a^2 > c^2$.

From Equation of Confocal Hyperbolas: Formulation 2:
 * $(1)$ defines a hyperbola when $a^2 < c^2$.

Thus it is seen that $(1)$ is that of a conic section.

We use the technique of formation of ordinary differential equation by elimination.

Differentiating $x$ gives:
 * $\dfrac {2 x} {a^2} + \dfrac {2 y} {a^2 - c^2} \dfrac {\d y} {\d x} = 0$

so
 * $\dfrac {\d y} {\d x} = - \dfrac {a^2 - c^2} {a^2} \dfrac x y$

Now we need to eliminate $c$ from the above.

We go back to $(1)$:

Substituting for $a^2 - c^2$:

This is separable, so separate it:
 * $\ds \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$

which leads to:

Now $k^2$ is still arbitrary at this point, so set:
 * $k^2 = \dfrac {a^2 - c^2} {a^2}$

Substituting this into the above gives us:
 * $\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$

which is $(1)$.

Hence the result by definition of self-orthogonal.