Q118 / 180 · KCET 2010 · Mathematics
$P$ 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
- A$\frac{1}{2 e}$
- B$\frac{1}{\mathrm{e}}$
- C$2 \sqrt{\mathrm{e}^{2}+1}$
- Correct answerD$\sqrt{\mathrm{e}^{2}+1}$
Solution
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 $=\left|\mathrm{e} \cdot 0+0-\left(\mathrm{e}^{2}+1\right)\right|$
$$
\begin{aligned}
&=\sqrt{e^{2}+1} \\
&=\frac{\left(e^{2}+1\right)}{\sqrt{e^{2}+1}} \\
&=\sqrt{e^{2}+1}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.