Confocal Conics are Self-Orthogonal

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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 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:

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


which leads to:

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

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.

$\blacksquare$


Sources

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