# 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.

## 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

$\blacksquare$