Slope of Tangent to Lemniscate 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

\(\displaystyle \paren {x^2 + y^2}^2\) \(=\) \(\displaystyle 2 a^2 \paren {x^2 - y^2}\) Definition of Lemniscate of Bernoulli
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\dfrac \d {\d x} } {\paren {x^2 + y^2}^2}\) \(=\) \(\displaystyle \map {\dfrac \d {\d x} } {2 a^2 \paren {x^2 - y^2} }\) differentiating with respect to $x$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \paren {x^2 + y^2} \paren {2 x + 2 y \dfrac {\d y} {\d x} }\) \(=\) \(\displaystyle 2 a^2 \paren {2 x - 2 y \dfrac {\d y} {\d x} }\) Derivative of Power and Chain Rule
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 y \paren {x^2 + y^2 + a^2} \dfrac {\d y} {\d x}\) \(=\) \(\displaystyle 4 x \paren {a^2 - \paren {x^2 + y^2} }\) gathering terms
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\d y} {\d x}\) \(=\) \(\displaystyle \dfrac {x \paren {a^2 - \paren {x^2 + y^2} } } {y \paren {x^2 + y^2 + a^2} }\) gathering terms



$\blacksquare$

Sources