Key Concepts and Formulas

• Implicit Differentiation: Used to find the derivative of a function where $y$ is not explicitly defined in terms of $x$.
• Slope of Tangent and Normal: The derivative $\frac{dy}{dx}$ gives the slope of the tangent. The slope of the normal is the negative reciprocal of the tangent's slope.
• Perpendicular Distance from a Point to a Line: The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Verify the point lies on the curve. Before proceeding, we confirm that the point (2,2) lies on the curve $x^2 + 2xy - 3y^2 = 0$. This is a good practice to avoid errors later. Substituting $x=2$ and $y=2$ into the equation: $(2)^2 + 2(2)(2) - 3(2)^2 = 4 + 8 - 12 = 0$ Since the equation holds true, the point (2,2) lies on the curve.

Step 2: Differentiate the equation implicitly. We need to find $\frac{dy}{dx}$ to determine the slope of the tangent. We differentiate both sides of the equation $x^2 + 2xy - 3y^2 = 0$ with respect to $x$, using the product rule and chain rule where necessary. $\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(3y^2) = \frac{d}{dx}(0)$ $2x + (2y + 2x\frac{dy}{dx}) - 6y\frac{dy}{dx} = 0$

Step 3: Solve for $\frac{dy}{dx}$. Now we isolate $\frac{dy}{dx}$ to find an expression for the slope of the tangent at any point on the curve. $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$ $2x\frac{dy}{dx} - 6y\frac{dy}{dx} = -2x - 2y$ $\frac{dy}{dx}(2x - 6y) = -2x - 2y$ $\frac{dy}{dx} = \frac{-2x - 2y}{2x - 6y} = \frac{-(x+y)}{x-3y} = \frac{x+y}{3y-x}$

Step 4: Evaluate $\frac{dy}{dx}$ at the point (2,2). We substitute $x=2$ and $y=2$ into the expression for $\frac{dy}{dx}$ to find the slope of the tangent at the point (2,2). $\frac{dy}{dx}\Big|_{(2,2)} = \frac{2+2}{3(2)-2} = \frac{4}{6-2} = \frac{4}{4} = 1$ So, the slope of the tangent at (2,2) is 1.

Step 5: Find the slope of the normal. The normal is perpendicular to the tangent. Therefore, the slope of the normal, $m_N$, is the negative reciprocal of the slope of the tangent. $m_N = -\frac{1}{1} = -1$

Step 6: Find the equation of the normal. We use the point-slope form of a line, $y - y_1 = m(x - x_1)$, with the point (2,2) and the slope of the normal, $m_N = -1$. $y - 2 = -1(x - 2)$ $y - 2 = -x + 2$ $x + y - 4 = 0$

Step 7: Calculate the perpendicular distance from the origin to the normal. The equation of the normal is $x + y - 4 = 0$. The origin is (0,0). We use the formula for the perpendicular distance from a point to a line: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ In this case, $A = 1$, $B = 1$, $C = -4$, $x_0 = 0$, and $y_0 = 0$. $d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs during implicit differentiation and when finding the negative reciprocal for the normal's slope.
• Simplifying Early: While tempting, simplifying the expression for $\frac{dy}{dx}$ before substituting the point can sometimes lead to more complex algebra.
• Forgetting the Chain Rule: Remember to apply the chain rule when differentiating terms involving $y$ with respect to $x$.

Summary

We found the equation of the normal to the curve at the point (2,2) using implicit differentiation and the concept of perpendicular slopes. Then, we calculated the perpendicular distance from the origin to this normal line using the appropriate formula. The perpendicular distance is $2\sqrt{2}$.

Final Answer The final answer is \boxed{2\sqrt{2}}, which corresponds to option (D).