Slope of Normal is Minus Reciprocal of Tangent

Theorem

Let $C$ be a curve defined by a real function which is differentiable.

Let $P$ be a point on $C$.

Let the curvature of $C$ at $P$ be non-zero.

Let $r$ be the slope of the tangent to $C$ at $P$.

Let $s$ be the slope of the normal to $C$ at $P$.

Then:

$r = -\dfrac 1 s$

Proof

By definition, the normal to $C$ at $P$ is defined as being perpendicular to the tangent at $P$ and in the same plane as $P$.

The result follows from Condition for Straight Lines in Plane to be Perpendicular.

$\blacksquare$