# 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. ## Proof

Consider:

$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
$(1)$ defines an ellipse when $a^2 > c^2$.
$(1)$ 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 with respect to $x$ 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 $(1)$:

 $\ds \frac {x^2} {a^2} + \frac {y^2} {a^2 - c^2}$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds \frac {y^2} {a^2 - c^2}$ $=$ $\ds 1 - \frac {x^2} {a^2}$ $\ds$ $=$ $\ds \frac {a^2 - x^2}{a^2}$ $\ds \leadsto \ \$ $\ds a^2 - c^2$ $=$ $\ds \frac {a^2 y^2} {a^2 - x^2}$

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

 $\ds \dfrac {\d y} {\d x}$ $=$ $\ds -\frac {a^2 y^2} {a^2 \paren {a^2 - x^2} } \frac x y$ $\ds$ $=$ $\ds \frac {x y} {a^2 - x^2}$

This is separable, so separate it:

$\ds \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$

 $\ds \ln y$ $=$ $\ds \frac 1 2 \, \map \ln {a^2 - x^2} + \ln k$ $\ds \leadsto \ \$ $\ds 2 \ln y$ $=$ $\ds \map \ln {a^2 - x^2} + 2 \ln k$ $\ds \leadsto \ \$ $\ds \ln y^2$ $=$ $\ds \map \ln {a^2 - x^2} + \ln k^2$ $\ds \leadsto \ \$ $\ds \ln y^2$ $=$ $\ds \map \ln {k^2 \paren {a^2 - x^2} }$ $\ds \leadsto \ \$ $\ds y^2$ $=$ $\ds k^2 \paren {a^2 - x^2}$ $\ds \leadsto \ \$ $\ds y^2 + k^2 x^2$ $=$ $\ds a^2$ $\ds \leadsto \ \$ $\ds \frac {x^2} {a^2} + \frac {y^2} {k^2 a^2}$ $=$ $\ds 1$

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.

$\blacksquare$