Differential Equation defining Confocal Conics

Theorem

Consider the equation:

$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$

where $a^2 > b^2$ and $-\lambda < a^2$.

defining the set of confocal conics whose foci are at $\tuple {\pm \sqrt {a^2 - b^2}, 0}$.

The differential equation defining these confocal conics is:

$x y \paren {\paren {y'}^2 - 1} + \paren {x^2 - y^2 - a^2 + b^2} y' = 0$

Proof

\(\ds \dfrac {x^2} {a^2 + \lambda}\)

\(=\)

\(\ds 1 - \dfrac {y^2} {b^2 + \lambda}\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \dfrac {b^2 + \lambda - y^2} {b^2 + \lambda}\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds a^2 + \lambda\)

\(=\)

\(\ds \dfrac {x^2 \paren {b^2 + \lambda} } {b^2 + \lambda - y^2}\)

\(\ds \dfrac {y^2} {b^2 + \lambda}\)

\(=\)

\(\ds 1 - \dfrac {x^2} {a^2 + \lambda}\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \dfrac {a^2 + \lambda - x^2} {a^2 + \lambda}\)

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds b^2 + \lambda\)

\(=\)

\(\ds \dfrac {y^2 \paren {a^2 + \lambda} } {a^2 + \lambda - x^2}\)

Then we have:

\(\ds \map {\dfrac \d {\d x} } {\dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} }\)

\(=\)

\(\ds \map {\dfrac \d {\d x} } 1\)

Differentiating $(1)$ with respect to $x$

\(\text {(4)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {2 x} {a^2 + \lambda} + \dfrac {2 y} {b^2 + \lambda} \dfrac {\d y} {\d x}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {a^2 + \lambda} y'\)

\(=\)

\(\ds -\dfrac {2 x} {2 y} \paren {b^2 + \lambda}\)

putting $y' = \dfrac {\d y} {\d x}$

\(\ds \)

\(=\)

\(\ds -\dfrac x y \dfrac {y^2 \paren {a^2 + \lambda} } {a^2 + \lambda - x^2}\)

from $(3)$

\(\ds \leadsto \ \ \)

\(\ds y'\)

\(=\)

\(\ds -\dfrac {x y} {a^2 + \lambda - x^2}\)

\(\ds \leadsto \ \ \)

\(\ds y' \paren {a^2 + \lambda}\)

\(=\)

\(\ds x^2 y' - x y\)

\(\text {(5)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds a^2 + \lambda\)

\(=\)

\(\ds \dfrac {x^2 y' - x y} {y'}\)

and similarly:

\(\ds \paren {b^2 + \lambda}\)

\(=\)

\(\ds -\dfrac {2 y} {2 x} \paren {a^2 + \lambda} y'\)

from $(4)$, putting $y' = \dfrac {\d y} {\d x}$

\(\ds \)

\(=\)

\(\ds -\dfrac {y y'} x \dfrac {x^2 \paren {b^2 + \lambda} } {b^2 + \lambda - y^2}\)

from $(2)$

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(=\)

\(\ds -\dfrac {x y y'} {b^2 + \lambda - y^2}\)

\(\text {(6)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds b^2 + \lambda\)

\(=\)

\(\ds y^2 - x y y'\)

Eliminating $\lambda$:

\(\ds b^2 + \paren {\dfrac {x^2 y' - x y} {y'} - a^2}\)

\(=\)

\(\ds y^2 - x y y'\)

from $(5)$ and $(6)$

\(\ds \leadsto \ \ \)

\(\ds y' \paren {b^2 - a^2} + x^2 y' - x y - y^2 y + x y \paren {y'}^2\)

\(=\)

\(\ds 0\)

multiplying through by $y'$ and rearranging

\(\ds \leadsto \ \ \)

\(\ds x y \paren {\paren {y'}^2 - 1} + \paren {x^2 - y^2 - a^2 + b^2} y'\)

\(=\)

\(\ds 0\)

multiplying through by $y'$ and rearranging

$\blacksquare$

Sources

• 1926: E.L. Ince: Ordinary Differential Equations ... (previous) ... (next): Chapter $\text I$: Introductory: $\S 1.201$ The Differential Equation of a Family of Confocal Conics