Key Concepts and Formulas

• Implicit Differentiation: Used to find the derivative $\frac{dy}{dx}$ when $y$ is not explicitly defined as a function of $x$. This derivative represents the slope of the tangent to the curve.
• Equation of a Tangent Line: Given a point $(x_1, y_1)$ on a curve and the slope $m$ of the tangent at that point, the equation of the tangent line is $y - y_1 = m(x - x_1)$.

Step-by-Step Solution

Step 1: Rewrite the given equation.

The given equation is $x^2 y^2 - 2x = 4(1 - y)$. Rewrite it as: $x^2y^2 - 2x = 4 - 4y$ $x^2y^2 - 2x + 4y - 4 = 0$

Step 2: Implicitly differentiate with respect to $x$.

Differentiate both sides of the equation $x^2y^2 - 2x + 4y - 4 = 0$ with respect to $x$: $\frac{d}{dx}(x^2y^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(4y) - \frac{d}{dx}(4) = 0$ Using the product rule and chain rule, we get: $(2x)(y^2) + (x^2)(2y\frac{dy}{dx}) - 2 + 4\frac{dy}{dx} = 0$ $2xy^2 + 2x^2y\frac{dy}{dx} - 2 + 4\frac{dy}{dx} = 0$

Step 3: Solve for $\frac{dy}{dx}$.

Rearrange the equation to isolate $\frac{dy}{dx}$: $2x^2y\frac{dy}{dx} + 4\frac{dy}{dx} = 2 - 2xy^2$ $\frac{dy}{dx}(2x^2y + 4) = 2 - 2xy^2$ $\frac{dy}{dx} = \frac{2 - 2xy^2}{2x^2y + 4}$ $\frac{dy}{dx} = \frac{1 - xy^2}{x^2y + 2}$

Step 4: Evaluate $\frac{dy}{dx}$ at the point (2, -2).

Substitute $x = 2$ and $y = -2$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx}\Big|_{(2, -2)} = \frac{1 - (2)(-2)^2}{(2)^2(-2) + 2} = \frac{1 - 2(4)}{4(-2) + 2} = \frac{1 - 8}{-8 + 2} = \frac{-7}{-6} = \frac{7}{6}$

Step 5: Find the equation of the tangent line.

Using the point-slope form of a line, the equation of the tangent line at $(2, -2)$ with slope $\frac{7}{6}$ is: $y - (-2) = \frac{7}{6}(x - 2)$ $y + 2 = \frac{7}{6}x - \frac{14}{6}$ $y = \frac{7}{6}x - \frac{7}{3} - 2$ $y = \frac{7}{6}x - \frac{7}{3} - \frac{6}{3}$ $y = \frac{7}{6}x - \frac{13}{3}$

Step 6: Check which point does NOT lie on the tangent line.

We need to find the point that does not satisfy the equation $y = \frac{7}{6}x - \frac{13}{3}$.

• **(A) $\left( {4,{1 \over 3}} \right)$: $\frac{1}{3} = \frac{7}{6}(4) - \frac{13}{3}$. $\frac{1}{3} = \frac{14}{3} - \frac{13}{3}$. $\frac{1}{3} = \frac{1}{3}$. This point lies on the line.
• (B) (8, 5): $5 = \frac{7}{6}(8) - \frac{13}{3}$. $5 = \frac{28}{3} - \frac{13}{3}$. $5 = \frac{15}{3}$. $5 = 5$. This point lies on the line.
• (C) (-4, -9): $-9 = \frac{7}{6}(-4) - \frac{13}{3}$. $-9 = -\frac{14}{3} - \frac{13}{3}$. $-9 = -\frac{27}{3}$. $-9 = -9$. This point lies on the line.
• (D) (-2, -7): $-7 = \frac{7}{6}(-2) - \frac{13}{3}$. $-7 = -\frac{7}{3} - \frac{13}{3}$. $-7 = -\frac{20}{3}$. This point does not lie on the line.

Therefore, the tangent line does not pass through the point $\left( {4,{1 \over 3}} \right)$.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs when differentiating and substituting. A single sign error can lead to an incorrect answer.
• Implicit Differentiation: Remember to apply the chain rule correctly when differentiating terms involving $y$ with respect to $x$. For example, $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$.
• Simplification: Simplify the expression for $\frac{dy}{dx}$ before substituting the point to avoid unnecessary calculations.

Summary

We found the equation of the tangent line to the curve $x^2y^2 - 2x = 4(1 - y)$ at the point (2, -2) using implicit differentiation. The equation of the tangent line is $y = \frac{7}{6}x - \frac{13}{3}$. By substituting the coordinates of each given point into the equation of the tangent line, we determined that the tangent line does not pass through the point $\left( {4,{1 \over 3}} \right)$.

Final Answer

The final answer is \boxed{\left( {4,{1 \over 3}} \right)}, which corresponds to option (A).