A) \[\frac{1}{2e}\]
B) \[\frac{1}{e}\]
C) \[2\sqrt{{{e}^{2}}+1}\]
D) \[\sqrt{{{e}^{2}}+1}\]
Correct Answer: D
Solution :
Given, curve \[y={{\log }_{e}}x\] ……(i) Let the coordinate of point of contact\[P(\alpha ,\beta )\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{x}\] 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 Eq. (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\[\frac{dy}{dx}=-x\] \[{{\left( \frac{dy}{dx} \right)}_{at\,(p)}}=-e\] Equation of normal at ‘P’ \[(y-1)=-e(x-e)\] \[y-1=-ex-{{e}^{2}}\] \[ex+y-({{e}^{2}}+1)=0\] ….(ii) The length of perpendicular drawn form the origin to the normal \[=|e.0+0-({{e}^{2}}+1)|\] \[=\sqrt{{{e}^{2}}+1}\] \[=\frac{({{e}^{2}}+1)}{\sqrt{{{e}^{2}}+1}}\] \[=\sqrt{{{e}^{2}}+1}\]
You need to login to perform this action.
You will be redirected in
3 sec
