## Curve for Which Magntitude of Curvature is equal to Length of Normal to x Axis

For a certain curve, the magnitude of the radius of curvature any point
$P(x_P,y_P)$
is equal to the length of the normal drawn from the point to the
$x$
axis.
In the diagram,
$| PQ | = | r | =\frac{(1+ (\frac{dy}{dx})^2)^{3/2}}{ \frac{d^2y}{dx^2}}$
.

The normal at
$(x_P,y_P)$
is
$y-y_P = -1/ \frac{dy}{dx}_{(x_P,y_P)} (x-x_P)$
. When
$y=0$
,
$y_P \frac{dy}{dx}_{(x_P,y_P)}=x_Q-x_P$
.
Hence
$| PQ | = \sqrt{y_P^2 +(x_Q-x_P)^2}=y_P \sqrt{1+ (\frac{dy}{dx}_{(x_P,y_P)})^2}$
.
$| PQ | = | r | \rightarrow y_P \sqrt{1+ (\frac{dy}{dx}_{(x_P,y_P)})^2} =\frac{(1+ (\frac{dy}{dx})^2)^{3/2}}{ \frac{d^2y}{dx^2}} \rightarrow y_P | \frac{d^2y}{dx^2}_{(x_P,y_P)} | = 1+(\frac{dy}{dx})^2)$
. Since
$P$
is a general point we can write
$y | \frac{d^2y}{dx^2}| = 1+(\frac{dy}{dx})^2)$
.
Suppose
$\frac{d^2y}{dx^2} \gt 0$
. Let
$\frac{dy}{dx}=p$
then can write the last equation as
$1+p^2=yp \frac{dp}{dy} \rightarrow \frac{1}{y}dy= \frac{p}{1+p^2} dp \rightarrow \int \frac{1}{y}dy= \int \frac{p}{1+p^2} dp \rightarrow ln(y)+c = \frac{1}{2} ln(1+p^2)$
.
This can be written as
$Ay^2=1+p^2= 1+ (\frac{dy}{dx})2 \frac{dy}{dx}=\sqrt{Ay^2-1}$
.
Separating variables and integrating again gives
$\frac{1}{\sqrt{A}} cos^{-1} y=x+B$
.
Similarly, letting
$\frac{d^2y}{dx^2} \lt 0$
and solving gives the equation of a family of circles.