\(P\) is the point of contact of the tangent form the origin to the curve \(y=\log _{e} x\). The length of the perpendicular drawn form the origin to the normal at \(P\) is

Given, curve \(\mathrm{y}=\log _{\mathrm{e}} \mathrm{x} \quad \text{...(i)}\)
Let the coordinate of point of contact \(P(\alpha, \beta)\) \(\Rightarrow\)
\[
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}}
\]
Now, equation of tangent at ' \(P\) '
\[
(y-\beta)=\frac{1}{\alpha}(x-\alpha)
\]
Since, the tangent passing through the origin ie, \((0,0)\)
\[
(0-\beta)=\frac{1}{\alpha}(0-\alpha) \Rightarrow \beta=1
\]
At 'P' from Eq. (i)
\[
\begin{aligned}
\beta &=\log _{\mathrm{e}} \alpha \\
\Rightarrow \quad 1 &=\log _{\mathrm{e}} \alpha
\end{aligned} \quad(\because \beta=1)
\]
\[
\Rightarrow \quad \log _{\mathrm{e}} \alpha=\log _{\mathrm{e}} \mathrm{e}
\]
\[
\alpha=\mathrm{e}
\]
So, point of contact is \(P(e, 1)\).
Now, slope of normal \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{x}\)
\[
\left(\frac{d y}{d x}\right)_{\text {at (P) }}=-e
\]
Equation of normal at ' \(\mathrm{P}\) '
\[
\begin{gathered}
(y-1)=-e(x-e) \\
y-1=-e x+e^{2} \\
e x+y-\left(e^{2}+1\right)=0 \quad \text{...(ii)}
\end{gathered}
\]
The length of perpendicular drawn from the origin to the normal \(=\left|\mathrm{e} \cdot 0+0-\left(\mathrm{e}^{2}+1\right)\right|\)
\[
\begin{aligned}
&=\sqrt{e^{2}+1} \\
&=\frac{\left(e^{2}+1\right)}{\sqrt{e^{2}+1}} \\
&=\sqrt{e^{2}+1}
\end{aligned}
\]