A) \[\frac{1}{\sqrt{5}}\]
B) \[-\frac{1}{\sqrt{5}}\]
C) \[\frac{2}{\sqrt{5}}\]
D) \[-\frac{2}{\sqrt{5}}\]
Correct Answer: C
Solution :
The equation of given curve is \[y={{e}^{2x}}={{x}^{2}}\] ...(i) On putting x = 0 in Eq. (i), we get y = 1 \[\therefore \]The given point is P (0,1). On differentiating Eq. (i) w.r.t. x, we get \[\frac{dy}{dx}=2{{e}^{2x}}+2x\] \[\therefore \]\[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}=2\] Equation of tangent at P(0,1) to Eq. (i) is \[y-1=2(x-0)\]\[\Rightarrow \]\[2x-y+1=0\] ?(ii) \[\therefore \]Required distance =length of perpendicular from (1,1) to the line \[2x-y+1=0\] \[=\frac{2-1+1}{\sqrt{4+1}}=\frac{2}{\sqrt{5}}\]
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