# Slope of Tangent to Lemniscate at Origin

Jump to navigation
Jump to search

## 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.

## 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 for Derivatives | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 4 y \paren {x^2 + y^2 + a^2} \dfrac {\d y} {\d x}\) | \(=\) | \(\displaystyle 4 x \paren {a^2 - \paren {x^2 + y^2} }\) | 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

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 11$: Special Plane Curves: Lemniscate: $11.3$