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Hint: The equation of the tangent at the point \[\left( {{x_1},{y_1}} \right)\] is \[x{x_1} - \left( {x + {x_1}} \right) + \dfrac{1}{2}\left( {y + {y_1}} \right) = 0\].

Substitute in this equation, and then use the given conditions to find the required value.

It is given that the equation of the curve is \[y = x\left( {2 - x} \right)\].

Rewriting the above equation, we get

\[ \Rightarrow y = 2x - {x^2}{\text{ ......eq.(1)}}\]

Here we have to find the equation of normal to the given curve.

We know that the normal and the tangent of a curve at a point on the curve are perpendicular, that is, \[{\text{Slope of normal}} = \dfrac{{ - 1}}{{{\text{Slope of tangent}}}}\].

Here to get the slope of the tangent lets differentiate the given curve with respect to \[x\].

Differentiating the eq.\[(1)\] with respect to \[x\], we get

\[

\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2x - {x^2}} \right) \\

\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}2x - \dfrac{d}{{dx}}{x^2} \\

\Rightarrow \dfrac{{dy}}{{dx}} = 2 - 2x \\

\]

Substituting the given point \[\left( {2,0} \right)\] in the above equation, we get

\[

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 2 - 2\left( 2 \right) \\

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 2 - 4 \\

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = - 2 \\

\]

So we have the slope of the tangent at the given point \[\left( {2,0} \right)\] is \[ - 2\].

We will now find the slope of normal using the above formula of slope of normal and slope of tangent to the curve.

\[

{\text{Slope of normal}} = \dfrac{{ - 1}}{{ - 2}} \\

= \dfrac{1}{2} \\

\]

Thus, the slope of the normal from the above equation at the given point is \[\dfrac{1}{2}\].

We know that the formula of the equation of a line is \[y - {y_1} = m\left( {x - {x_1}} \right)\], where \[m\] is the slope of the line passing through the point \[\left( {{x_1},{y_1}} \right)\].

Substituting the value of slope of line in the above formula of equation of line passing through the given point \[\left( {2,0} \right)\], we get

\[

\Rightarrow y - 0 = \dfrac{1}{2}\left( {x - 2} \right) \\

\Rightarrow 2y = x - 2 \\

\Rightarrow x - 2y = 2 \\

\]

Thus, we have the equation of the normal to the curve is \[x - 2y = 2\].

Hence, the option A is correct.

Note: In solving these types of questions, you should be familiar with the concept of the equation of the tangent and normal to the curve and formula to find the slope. We should note the point that the normal of the given curve will pass through the given point and also we have to use the relation between normal and tangent to solve this question easily. Also, we are supposed to write the values properly to avoid any miscalculation.

Substitute in this equation, and then use the given conditions to find the required value.

__Complete step-by-step solution__It is given that the equation of the curve is \[y = x\left( {2 - x} \right)\].

Rewriting the above equation, we get

\[ \Rightarrow y = 2x - {x^2}{\text{ ......eq.(1)}}\]

Here we have to find the equation of normal to the given curve.

We know that the normal and the tangent of a curve at a point on the curve are perpendicular, that is, \[{\text{Slope of normal}} = \dfrac{{ - 1}}{{{\text{Slope of tangent}}}}\].

Here to get the slope of the tangent lets differentiate the given curve with respect to \[x\].

Differentiating the eq.\[(1)\] with respect to \[x\], we get

\[

\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2x - {x^2}} \right) \\

\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}2x - \dfrac{d}{{dx}}{x^2} \\

\Rightarrow \dfrac{{dy}}{{dx}} = 2 - 2x \\

\]

Substituting the given point \[\left( {2,0} \right)\] in the above equation, we get

\[

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 2 - 2\left( 2 \right) \\

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = 2 - 4 \\

\Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 2}} = - 2 \\

\]

So we have the slope of the tangent at the given point \[\left( {2,0} \right)\] is \[ - 2\].

We will now find the slope of normal using the above formula of slope of normal and slope of tangent to the curve.

\[

{\text{Slope of normal}} = \dfrac{{ - 1}}{{ - 2}} \\

= \dfrac{1}{2} \\

\]

Thus, the slope of the normal from the above equation at the given point is \[\dfrac{1}{2}\].

We know that the formula of the equation of a line is \[y - {y_1} = m\left( {x - {x_1}} \right)\], where \[m\] is the slope of the line passing through the point \[\left( {{x_1},{y_1}} \right)\].

Substituting the value of slope of line in the above formula of equation of line passing through the given point \[\left( {2,0} \right)\], we get

\[

\Rightarrow y - 0 = \dfrac{1}{2}\left( {x - 2} \right) \\

\Rightarrow 2y = x - 2 \\

\Rightarrow x - 2y = 2 \\

\]

Thus, we have the equation of the normal to the curve is \[x - 2y = 2\].

Hence, the option A is correct.

Note: In solving these types of questions, you should be familiar with the concept of the equation of the tangent and normal to the curve and formula to find the slope. We should note the point that the normal of the given curve will pass through the given point and also we have to use the relation between normal and tangent to solve this question easily. Also, we are supposed to write the values properly to avoid any miscalculation.

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