Slope of Lemniscate of Bernoulli at Origin

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Theorem

Consider the lemniscate of Bernoulli $M$ embedded in a Cartesian plane such that its foci are at $\tuple {a, 0}$ and $\tuple {-a, 0}$ respectively.

Let $O$ denote the origin.


The tangents to $M$ at $O$ are at an angle of $45 \degrees = \dfrac \pi 4$ to the $x$-axis.


Lemniscate-tangents-at-origin.png


Proof

Recall the parametric definition of lemniscate of Bernoulli:

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

Note that $\tuple {x, y} = \tuple {0, 0}$ if and only if $\cos t = 0$.

At these points we have $\sin t = \pm \sqrt {1 - \cos^2 t} = \pm 1$.


We have:

\(\ds \frac {\d x} {\d t}\) \(=\) \(\ds a \sqrt 2 \paren {\frac {-\sin t \paren {\sin^2 t + 1} - \cos t \paren {2 \sin t \cos t} } {\paren {\sin^2 t + 1}^2} }\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {-\sin^3 t - 2 \sin t \cos^2 t - \sin t}\)
\(\ds \frac {\d y} {\d t}\) \(=\) \(\ds a \sqrt 2 \paren {\frac {\paren {\cos^2 t - \sin^2 t} \paren {\sin^2 t + 1} - \cos t \sin t \paren {2 \sin t \cos t} } {\paren {\sin^2 t + 1}^2} }\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {\cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t - 2 \cos^2 t \sin^2 t}\)
\(\ds \) \(=\) \(\ds \frac {a \sqrt 2} {\paren {\sin^2 t + 1}^2} \paren {- \cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {\d y} {\d t} \times \frac {\d t} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {-\cos^2 t \sin^2 t - \sin^4 t + \cos^2 t - \sin^2 t} {-\sin^3 t - 2 \sin t \cos^2 t - \sin t}\)

By substituting $\cos t = 0$ and $\sin t = \pm 1$, we have:

\(\ds \valueat {\frac {\d y} {\d x} } {\tuple {x, y} \mathop = \tuple {0, 0} }\) \(=\) \(\ds \frac {0 - \paren {\pm 1}^4 + 0 - \paren {\pm 1}^2} {-\paren {\pm 1}^3 - 0 - \paren {\pm 1} }\)
\(\ds \) \(=\) \(\ds \frac {-2} {\mp 2}\)
\(\ds \) \(=\) \(\ds \pm 1\)

As $\map \tan {\pm 45 \degrees} = \pm 1$, these tangents at $O$ are at an angle of $45 \degrees$ to the $x$-axis.

$\blacksquare$


Sources

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