What is the distance from the origin of the normal to the curve $x = 2\cos t + 2t\sin t$, $y = 2\sin t - 2t\cos t$ at $t = \dfrac{\pi }{4}$?
$
  {\text{A}}{\text{. 2}} \\
  {\text{B}}{\text{. 4}} \\
  {\text{C}}{\text{. }}\sqrt {\text{2}} \\
  {\text{D}}{\text{. 2}}\sqrt {\text{2}} \\
 $

Hint- Here, we will be proceeding by differentiating both the given parametric equations of the curve with respect to t. Then, we will use the formula i.e., Slope of the normal to any curve is given by $ - \dfrac{{dx}}{{dy}}$.

Complete Step-by-Step solution:
Given equations of the curve in parametric form are $x = 2\cos t + 2t\sin t{\text{ }} \to {\text{(1)}}$ and $y = 2\sin t - 2t\cos t{\text{ }} \to {\text{(2)}}$
By differentiating equation (1) both sides with respect to t, we get
\[
   \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {2\cos t + 2t\sin t} \right) \\
   \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {2\cos t} \right) + \dfrac{d}{{dt}}\left( {2t\sin t} \right) \\
   \Rightarrow \dfrac{{dx}}{{dt}} = 2\dfrac{d}{{dt}}\left( {\cos t} \right) + 2\dfrac{d}{{dt}}\left( {t\sin t} \right) \\
   \Rightarrow \dfrac{{dx}}{{dt}} = 2\left( { - \sin t} \right) + 2\left( {t\cos t + \sin t} \right) \\
   \Rightarrow \dfrac{{dx}}{{dt}} = - 2\sin t + 2t\cos t + 2\sin t \\
   \Rightarrow \dfrac{{dx}}{{dt}} = 2t\cos t{\text{ }} \to {\text{(3)}} \\
 \]
By differentiating equation (2) both sides with respect to t, we get
\[
   \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {2\sin t - 2t\cos t} \right) \\
   \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {2\sin t} \right) - \dfrac{d}{{dt}}\left( {2t\cos t} \right) \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 2\dfrac{d}{{dt}}\left( {\sin t} \right) - 2\dfrac{d}{{dt}}\left( {t\cos t} \right) \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 2\left( {\cos t} \right) - 2\left[ {t\left( { - \sin t} \right) + \cos t} \right] \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 2\cos t - 2\left[ { - t\sin t + \cos t} \right] \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 2\cos t + 2t\sin t - 2\cos t \\
   \Rightarrow \dfrac{{dy}}{{dt}} = 2t\sin t{\text{ }} \to {\text{(4)}} \\
 \]
By dividing equation (3) by equation (4), we get
\[
   \Rightarrow \dfrac{{\left( {\dfrac{{dx}}{{dt}}} \right)}}{{\left( {\dfrac{{dy}}{{dt}}} \right)}} = \dfrac{{2t\cos t}}{{2t\sin t}} \\
   \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{{\cos t}}{{\sin t}} \\
 \]
According to the definition of cotangent trigonometric function, \[\cot t = \dfrac{{\cos t}}{{\sin t}}\]
\[ \Rightarrow \dfrac{{dx}}{{dy}} = \cot t\]
As we know that the slope of the normal to any curve is given by $ - \dfrac{{dx}}{{dy}}$
Since, $\cot \left( {\dfrac{\pi }{4}} \right) = 1$
Slope of the normal to the given curve at $t = \dfrac{\pi }{4}$ = $ - \dfrac{{dx}}{{dy}} = - \cot t = - \cot \left( {\dfrac{\pi }{4}} \right) = - 1$
Put $t = \dfrac{\pi }{4}$in equations (1) and (2), we get
$ \Rightarrow x = 2\cos \left( {\dfrac{\pi }{4}} \right) + 2\left( {\dfrac{\pi }{4}} \right)\sin \left( {\dfrac{\pi }{4}} \right)$
Using $\sin \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ in the above equation, we get
$
   \Rightarrow x = 2\left( {\dfrac{1}{{\sqrt 2 }}} \right) + 2\left( {\dfrac{\pi }{4}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow x = \dfrac{2}{{\sqrt 2 }} + \dfrac{\pi }{{2\sqrt 2 }} \\
   \Rightarrow x = \dfrac{{\left( {2 \times 2} \right) + \pi }}{{2\sqrt 2 }} \\
   \Rightarrow x = \dfrac{{4 + \pi }}{{2\sqrt 2 }} \\
 $
Also, $
   \Rightarrow y = 2\sin \left( {\dfrac{\pi }{4}} \right) - 2\left( {\dfrac{\pi }{4}} \right)\cos \left( {\dfrac{\pi }{4}} \right) \\
   \Rightarrow y = 2\left( {\dfrac{1}{{\sqrt 2 }}} \right) - 2\left( {\dfrac{\pi }{4}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow y = \dfrac{2}{{\sqrt 2 }} - \dfrac{\pi }{{2\sqrt 2 }} \\
   \Rightarrow y = \dfrac{{\left( {2 \times 2} \right) - \pi }}{{2\sqrt 2 }} \\
   \Rightarrow y = \dfrac{{4 - \pi }}{{2\sqrt 2 }} \\
 $
So, the normal to the given curve must be passing through the point ($\dfrac{{4 + \pi }}{{2\sqrt 2 }}$,$\dfrac{{4 - \pi }}{{2\sqrt 2 }}$)
As we know that the equation of any straight line having slope m and passing through point (a,b) is given by y-b = m(x-a)
So, the equation of the normal having slope of -1 and passing through the point ($\dfrac{{4 + \pi }}{{2\sqrt 2 }}$,$\dfrac{{4 - \pi }}{{2\sqrt 2 }}$) is given by
\[
   \Rightarrow y - \left( {\dfrac{{4 - \pi }}{{2\sqrt 2 }}} \right) = - 1\left[ {x - \left( {\dfrac{{4 + \pi }}{{2\sqrt 2 }}} \right)} \right] \\
   \Rightarrow y - \left( {\dfrac{{4 - \pi }}{{2\sqrt 2 }}} \right) = \left( {\dfrac{{4 + \pi }}{{2\sqrt 2 }}} \right) - x \\
   \Rightarrow x + y - \left( {\dfrac{{4 - \pi }}{{2\sqrt 2 }}} \right) - \left( {\dfrac{{4 + \pi }}{{2\sqrt 2 }}} \right) = 0 \\
   \Rightarrow x + y - \left[ {\dfrac{{\left( {4 - \pi } \right) + \left( {4 + \pi } \right)}}{{2\sqrt 2 }}} \right] = 0 \\
   \Rightarrow x + y - \left[ {\dfrac{{4 - \pi + 4 + \pi }}{{2\sqrt 2 }}} \right] = 0 \\
   \Rightarrow x + y - \left[ {\dfrac{8}{{2\sqrt 2 }}} \right] = 0 \\
   \Rightarrow x + y + \left( { - \dfrac{4}{{\sqrt 2 }}} \right) = 0 \\
 \]
Since, the distance from point $\left( {{x_1},{y_1}} \right)$ of the straight line $ax + by +c = 0$ is given by
$d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
So, the distance from origin (0,0) of the normal \[x + y + \left( { - \dfrac{4}{{\sqrt 2 }}} \right) = 0\] is given by
$
   \Rightarrow d = \left| {\dfrac{{\left( {1 \times 0} \right) + \left( {1 \times 0} \right) + \left( { - \dfrac{4}{{\sqrt 2 }}} \right)}}{{\sqrt {{1^2} + {1^2}} }}} \right| \\
   \Rightarrow d = \left| {\dfrac{{0 + 0 - \left( {\dfrac{4}{{\sqrt 2 }}} \right)}}{{\sqrt 2 }}} \right| \\
   \Rightarrow d = \left| {\dfrac{{ - \dfrac{4}{{\sqrt 2 }}}}{{\sqrt 2 }}} \right| \\
   \Rightarrow d = \left| {\dfrac{{ - 4}}{{\sqrt 2 \times \sqrt 2 }}} \right| \\
   \Rightarrow d = \left| {\dfrac{{ - 4}}{2}} \right| = \left| { - 2} \right| \\
   \Rightarrow d = 2 \\
 $
Therefore, the distance from origin (0,0) of the normal to the curve $x = 2\cos t + 2t\sin t$, $y = 2\sin t - 2t\cos t$ at $t = \dfrac{\pi }{4}$ is 2.
Hence, option A is correct.

Note- In this particular problem, we have compared the equation of the normal to the given curve i.e., \[x + y + \left( { - \dfrac{4}{{\sqrt 2 }}} \right) = 0\] with the general equation of the straight line i.e., $ax + by = c$ we will get a = 1, y = 1 and c = \[ - \dfrac{4}{{\sqrt 2 }}\]. Also, here the point through which the normal to the given curve is passing is located at $t = \dfrac{\pi }{4}$.