Question

$P$ is the point of contact of the tangent from the origin to the curve $y = log_ex$ The length of the perpendicular drawn from the origin to the normal at $P$ is ______

• A. $2\sqrt {e^2+1}$
• B. $\sqrt {e^2+1}$
• C. $\frac {1}{2e}$
• D. $\frac {1}{e}$

Answer 1

The Correct Option is B

Given, curve $y=\log _{e} x$...(i)
Let the coordinate of point of contact $P(\alpha, \beta)$
$\Rightarrow \frac{d y}{d x}= \frac{1}{x}$
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 E (i)
$\beta =\log _{e} \alpha$
$\Rightarrow 1 =\log _{e} \alpha$
$(\because \beta=1)$
$\Rightarrow \log _{e} \alpha =\log _{e} e$
$\alpha =e$
So, point of contact is $P(e, 1)$.
Now, slope of normal $\frac{d y}{d x}=-x$
$\left(\frac{d y}{d x}\right)_{ at (P)}=-e$
Equation of normal at 'P'
$(y-1)=-e(x-e)$
$y-1=-e x+ e^{2}$
$e x +y-\left(e^{2}+1\right)=0$...(ii)
The length of perpendicular drawn from the origin to the normal
$=\left|e \cdot 0+0-\left(e^{2}+1\right)\right|$
$=\sqrt{e^{2}+1}$
$=\frac{\left(e^{2}+1\right)}{\sqrt{e^{2}+1}}$
$=\sqrt{e^{2}+1}$