Equivalence of Definitions of Lemniscate of Bernoulli

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Theorem

The following definitions of the concept of Lemniscate of Bernoulli are equivalent:

Geometric Definition

Let $P_1$ and $P_2$ be points in the plane such that $P_1 P_2 = 2 a$ for some constant $a$.

The lemniscate of Bernoulli is the locus of points $M$ in the plane such that:

$P_1 M \times P_2 M = a^2$


Cartesian Definition

The lemniscate of Bernoulli is the curve defined by the Cartesian equation:

$\paren {x^2 + y^2}^2 = 2 a^2 \paren {x^2 - y^2}$


Polar Definition

The lemniscate of Bernoulli is the curve defined by the polar equation:

$r^2 = 2 a^2 \cos 2 \theta$


Parametric Definition

The lemniscate of Bernoulli is the curve defined by the parametric equation:

$\begin{cases} x = \dfrac {a \sqrt 2 \cos t} {\sin^2 t + 1} \\ y = \dfrac {a \sqrt 2 \cos t \sin t} {\sin^2 t + 1} \end{cases}$


Proof

Geometric Definition equivalent to Cartesian Definition

Let $M$ be a lemniscate of Bernoulli by the geometric definition.

Then by definition:

Let $P_1$ and $P_2$ be points in the plane such that $P_1 P_2 = 2 a$ for some constant $a$.

The lemniscate of Bernoulli is the locus of points $M$ in the plane such that:

$P_1 M \times P_2 M = a^2$


Let $P_1 = \tuple {a, 0}$ and $P_2 = \tuple {-a, 0}$.

Let $p = \tuple {x, y}$ be an arbitrary point of $M$.

We have:

\(\ds P_1 p\) \(=\) \(\ds \sqrt {\paren {a - x}^2 + y^2}\) Distance Formula
\(\ds \) \(=\) \(\ds \sqrt {a^2 - 2 a x + x^2 + y^2}\)
\(\ds P_2 p\) \(=\) \(\ds \sqrt {\paren {a + x}^2 + y^2}\) Distance Formula
\(\ds \) \(=\) \(\ds \sqrt {a^2 + 2 a x + x^2 + y^2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {P_1 p} \paren {P_2 p}\) \(=\) \(\ds \sqrt {a^2 - 2 a x + x^2 + y^2} \sqrt {a^2 + 2 a x + x^2 + y^2}\) Definition of Lemniscate of Bernoulli
\(\ds \) \(=\) \(\ds a^2\)
\(\ds \leadsto \ \ \) \(\ds a^4\) \(=\) \(\ds \paren {a^2 - 2 a x + x^2 + y^2} \paren {a^2 + 2 a x + x^2 + y^2}\)
\(\ds \) \(=\) \(\ds \paren {a^2 + \paren {x^2 + y^2} }^2 - 4 a^2 x^2\) Difference of Two Squares
\(\ds \) \(=\) \(\ds a^4 + 2 a^2 \paren {x^2 + y^2} + \paren {x^2 + y^2}^2 - 4 a^2 x^2\) Square of Sum
\(\ds \) \(=\) \(\ds a^4 + 2 a^2 \paren {y^2 - x^2} + \paren {x^2 + y^2}^2\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {x^2 + y^2}^2\) \(=\) \(\ds 2 a^2 \paren {x^2 - y^2}\)

Thus $M$ is a lemniscate of Bernoulli by the Cartesian definition.

$\Box$


Geometric Definition equivalent to Polar Definition

Let $M$ be a lemniscate of Bernoulli by the geometric definition.

Then by definition:

Let $P_1$ and $P_2$ be points in the plane such that $P_1 P_2 = 2 a$ for some constant $a$.

The lemniscate of Bernoulli is the locus of points $M$ in the plane such that:

$P_1 M \times P_2 M = a^2$


Let $M$ be embedded in a polar coordinate plane whose origin is at $O$ and such that $P_1 = \polar {a, 0}$ and $P_2 = \polar {a, \pi}$.


Lemniscate-polar-construction.png


Consider an arbitrary point $p = \polar {r, \theta}$.

Let $d_1 = \size {P_1 p}$ and $d_2 = \size {P_2 p}$.

We have:

\(\ds d_1 d_2\) \(=\) \(\ds a^2\) Definition of Lemniscate of Bernoulli
\(\ds \) \(=\) \(\ds \sqrt {r^2 + a^2 - 2 a r \cos \theta} \times \sqrt {r^2 + a^2 - 2 a r \, \map \cos {\pi - \theta} }\) Cosine Rule
\(\ds \leadsto \ \ \) \(\ds \paren {a^2}^2\) \(=\) \(\ds \paren {r^2 + a^2 - 2 a r \cos \theta} \paren {r^2 + a^2 + 2 a r \cos \theta}\) Cosine of Supplementary Angle, and squaring throughout
\(\ds \) \(=\) \(\ds \paren {r^2 + a^2}^2 - \paren {2 a r \cos \theta}^2\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {r^2}^2 + 2 a^2 r^2 + \paren {a^2}^2 - 4 a^2 r^2 \cos^2 \theta\)
\(\ds \leadsto \ \ \) \(\ds r^2\) \(=\) \(\ds 2 a^2 \paren {2 \cos^2 \theta - 1}\) simplifying and rearranging
\(\ds \leadsto \ \ \) \(\ds r^2\) \(=\) \(\ds 2 a^2 \cos 2 \theta\) Double Angle Formula for Cosine: Corollary $1$

$\Box$


Parametric Definition equivalent to Cartesian Definition

Let $M$ be a lemniscate of Bernoulli by the parametric definition.

Then by definition:

The lemniscate of Bernoulli is the curve defined by the parametric equation:

$\begin{cases} x = \dfrac {a \sqrt 2 \cos t} {\sin^2 t + 1} \\ y = \dfrac {a \sqrt 2 \cos t \sin t} {\sin^2 t + 1} \end{cases}$


We have:

\(\ds x^2 - y^2\) \(=\) \(\ds \dfrac {2 a^2 \cos^2 t} {\paren {\sin^2 t + 1}^2} - \dfrac {2 a^2 \cos^2 t \sin^2 t} {\paren {\sin^2 t + 1}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {2 a^2 \cos^2 t \paren {1 - \sin^2 t} } {\paren {\sin^2 t + 1}^2}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {2 a^2 \paren {1 - \sin^2 t}^2} {\paren {\sin^2 t + 1}^2}\) Sum of Squares of Sine and Cosine


Then:

\(\ds x^2 + y^2\) \(=\) \(\ds \dfrac {2 a^2 \cos^2 t} {\paren {\sin^2 t + 1}^2} + \dfrac {2 a^2 \cos^2 t \sin^2 t} {\paren {\sin^2 t + 1}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {2 a^2 \cos^2 t \paren {1 + \sin^2 t} } {\paren {\sin^2 t + 1}^2}\)
\(\ds \) \(=\) \(\ds \dfrac {2 a^2 \paren {1 - \sin^2 t} \paren {1 + \sin^2 t} } {\paren {\sin^2 t + 1}^2}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \dfrac {2 a^2 \paren {1 - \sin^2 t} } {\sin^2 t + 1}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {x^2 + y^2}^2\) \(=\) \(\ds 2 a^2 \dfrac {2 a^2 \paren {1 - \sin^2 t}^2 } {\paren {\sin^2 t + 1}^2}\) squaring both sides and extracting $2 a^2$
\(\ds \) \(=\) \(\ds 2 a^2 \paren {x^2 - y^2}\) from $(1)$

$\blacksquare$

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