# Question #ec325

Jul 28, 2016

The perpendicular from origin $O$ to a normal to $y = {\log}_{e} \left(x\right)$ at point $P$ is the segment $O P$.
It's length is $\sqrt{{X}_{P}^{2} + {Y}_{P}^{2}}$, where ${X}_{P}$ is the abscissa of point $P$ and ${Y}_{P} = {\log}_{e} \left({X}_{P}\right)$.

#### Explanation:

Normal to a curve at point $P$ lying on this curve, by definition, is a line perpendicular to a tangent to a curve at this point (presuming the curve is smooth and has a tangent, otherwise a normal is undefined).

Since $O P$ is a tangent to curve $y = {\log}_{e} \left(x\right)$ from origin $O$ to point $P$ lying on this curve, it is also a perpendicular to a normal to a curve at point $P$.

So, our task is to measure the length of $O P$.
This is done by Pythagorean Theorem as the distance between two points:
origin $O$ with coordinates $\left(0 , 0\right)$ and
point $P$ on a curve with coordinates $\left({X}_{P} , {Y}_{P}\right)$, where
${Y}_{P} = {\log}_{e} \left({X}_{P}\right)$.

$O P = \sqrt{{X}_{P}^{2} + {Y}_{P}^{2}} = \sqrt{{X}_{P}^{2} + {\log}_{e}^{2} \left({X}_{P}\right)}$