The normal to the curve \[y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6\] at the point where the curve intersects the y-axis passes through the point.
A.\[\left( { - \dfrac{1}{2}, - \dfrac{1}{2}} \right)\]
B.\[\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)\] \[\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)\]
C.\[\left( {\dfrac{1}{2}, - \dfrac{1}{3}} \right)\]
D.\[\left( {\dfrac{1}{2},\dfrac{1}{3}} \right)\]

Hint: We simply need to find the equation of the normal which passes through the intersection point of the curve \[y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6\] and y-axis. We can then check which point as mentioned in the options satisfies the equation.

Complete step-by-step answer:
The equation of the curve is given by \[y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6\] and it intersects the y-axis.
Let us find the point of intersection of the given curve and the y-axis. Let this point of intersection is \[\left( {0,\alpha } \right)\]. Since \[\left( {0,\alpha } \right)\] lies on the given curve, hence it will satisfy the equation of the curve.
Therefore,
\[
  \alpha \left( {0 - 2} \right)\left( {0 - 3} \right) = 0 + 6 \\
   \Rightarrow \alpha \left( { - 2} \right)\left( { - 3} \right) = 6 \\
   \Rightarrow \alpha = 1 \\
 \]
Hence, the curve intersects the y-axis at point\[\left( {0,1} \right)\].
Now, we need to find the equation of normal to curve at point. We first find the slope of tangent at point\[\left( {0,1} \right)\].
\[
  \because y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6 \\
   \Rightarrow y\left( {{x^2} - 5x + 6} \right) = x + 6 \\
 \]
Differentiating both sides with respect to x, we get \[\left( {0,1} \right)\] is \[1\]. We can now use slope point form to find the equation of the normal.
Putting \[y = 1\] and \[x = 0\] in above equation, we have
\[
  1\left( {0 - 5} \right) + \left( {0 - 0 + 6} \right){\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\
   \Rightarrow - 5 + 6{\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\
   \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = \dfrac{6}{6} \\
  \therefore {\left( {\dfrac{{dy}}{{dx}}} \right)_{0,1}} = 1 \\
 \]
 Slope of tangent X slope of normal=\[ - 1\]
 Slope of normal\[ = - \dfrac{1}{{{\text{slope of tangent}}}}\]
\[
   = \dfrac{{ - 1}}{{ - 1}} \\
   = 1 \\
 \]
Thus the slope of normal which passes through \[\left( {0,1} \right)\] is\[1\]. We can now use slope- point form to find the equation of the normal.
\[
  \dfrac{{y - 1}}{{x - 0}} = 1 \\
   \Rightarrow y + x = 1 \\
 \]
Clearly, only option (c) satisfies the above equation of normal
So, the correct answer is “Option C”.

Note: The student must keep in mind that the x-coordinate of any point on y-axis is always 0. It is also very important to remember the relation between the slopes of tangent and normal at a point. Any point which lies on a curve must satisfy the equation of the curve.