Key Concepts and Formulas

• Equation of a Normal: The equation of the normal to a curve $y = f(x)$ at a point $(x_1, y_1)$ is given by $y - y_1 = m_N(x - x_1)$, where $m_N$ is the slope of the normal.
• Slope of Tangent and Normal: The slope of the tangent at a point is given by the derivative $\frac{dy}{dx}$ evaluated at that point. The slope of the normal is the negative reciprocal of the tangent's slope, i.e., $m_N = -\frac{1}{m_T}$, where $m_T$ is the slope of the tangent.
• Finding y-intercept: The y-intercept of a curve is the point where the curve intersects the y-axis, which occurs when $x = 0$.

Step-by-Step Solution

Step 1: Find the point of intersection with the y-axis.

The point where the curve intersects the y-axis has an x-coordinate of 0. We substitute $x = 0$ into the given equation to find the corresponding y-coordinate.

The equation of the curve is: $y(x - 2)(x - 3) = x + 6$ Substituting $x = 0$: $y(0 - 2)(0 - 3) = 0 + 6$ $y(-2)(-3) = 6$ $6y = 6$ $y = 1$ Thus, the curve intersects the y-axis at the point $(0, 1)$.

Step 2: Implicitly differentiate the equation to find $\frac{dy}{dx}$.

We need to find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent. Since it's difficult to isolate $y$, we'll use implicit differentiation.

The equation is: $y(x - 2)(x - 3) = x + 6$ $y(x^2 - 5x + 6) = x + 6$ Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{dy}{dx}(x^2 - 5x + 6) + y(2x - 5) = 1$

Step 3: Evaluate $\frac{dy}{dx}$ at the point (0, 1).

We substitute $x = 0$ and $y = 1$ into the differentiated equation to find the slope of the tangent at the point (0, 1).

$\frac{dy}{dx}(0^2 - 5(0) + 6) + 1(2(0) - 5) = 1$ $\frac{dy}{dx}(6) - 5 = 1$ $6\frac{dy}{dx} = 6$ $\frac{dy}{dx} = 1$ So, the slope of the tangent at the point (0, 1) is $m_T = 1$.

Step 4: Find the slope of the normal.

The slope of the normal is the negative reciprocal of the slope of the tangent.

$m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1$ The slope of the normal is $-1$.

Step 5: Find the equation of the normal.

We use the point-slope form of a line to find the equation of the normal. The point is $(0, 1)$ and the slope is $m_N = -1$.

$y - y_1 = m_N(x - x_1)$ $y - 1 = -1(x - 0)$ $y - 1 = -x$ $x + y = 1$

Step 6: Check which of the given options satisfies the equation of the normal.

We substitute the coordinates of each option into the equation $x + y = 1$ to see which point lies on the normal.

(A) $\left( \frac{1}{2}, \frac{1}{2} \right)$: $\frac{1}{2} + \frac{1}{2} = 1$. This is true. (B) $\left( \frac{1}{2}, - \frac{1}{3} \right)$: $\frac{1}{2} - \frac{1}{3} = \frac{1}{6} \neq 1$. This is false. (C) $\left( \frac{1}{2}, \frac{1}{3} \right)$: $\frac{1}{2} + \frac{1}{3} = \frac{5}{6} \neq 1$. This is false. (D) $\left( - \frac{1}{2}, - \frac{1}{3} \right)$: $- \frac{1}{2} - \frac{1}{3} = - \frac{5}{6} \neq 1$. This is false.

Only option (A) satisfies the equation of the normal.

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when differentiating and substituting values.
• Tangent vs. Normal: Remember to take the negative reciprocal of the tangent's slope to find the normal's slope.
• Implicit Differentiation: Be careful when applying the product rule and chain rule during implicit differentiation.

Summary

We found the point of intersection of the curve with the y-axis, implicitly differentiated the equation to find the slope of the tangent, calculated the slope of the normal, and then found the equation of the normal. Finally, we checked which of the given points satisfied the equation of the normal.

The final answer is $\boxed{\left( {{1 \over 2},{1 \over 2}} \right)}$, which corresponds to option (A).