Slope of Tangent to Lemniscate at Origin

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 $\tuple {x, y} = \tuple {0, 0}$ $\cos t = 0$.

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

We have:

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

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