Key Concepts and Formulas

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ when $y$ is not explicitly defined as a function of $x$.
• Normal Line: A line perpendicular to the tangent at a point on a curve. The slope of the normal ($m_n$) is the negative reciprocal of the slope of the tangent ($m_t$), i.e., $m_n = -\frac{1}{m_t}$.
• Point-Slope Form of a Line: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

Step-by-Step Solution

Step 1: Implicit Differentiation

We are given the equation of the curve: $x^2 + 2xy - 3y^2 = 0$ We need to find $\frac{dy}{dx}$ using implicit differentiation. Differentiating both sides with respect to $x$, we get: $\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(3y^2) = \frac{d}{dx}(0)$ $2x + 2\left(x\frac{dy}{dx} + y\frac{d}{dx}(x)\right) - 3\left(2y\frac{dy}{dx}\right) = 0$ $2x + 2x\frac{dy}{dx} + 2y - 6y\frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Now, we isolate $\frac{dy}{dx}$: $2x\frac{dy}{dx} - 6y\frac{dy}{dx} = -2x - 2y$ $\frac{dy}{dx}(2x - 6y) = -2x - 2y$ $\frac{dy}{dx} = \frac{-2x - 2y}{2x - 6y} = \frac{-x - y}{x - 3y} = \frac{x + y}{3y - x}$

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

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

Step 4: Find the Slope of the Normal

The slope of the normal ($m_n$) is the negative reciprocal of the slope of the tangent: $m_n = -\frac{1}{1} = -1$

Step 5: Find the Equation of the Normal

Using the point-slope form of a line, the equation of the normal at (1, 1) is: $y - 1 = -1(x - 1)$ $y - 1 = -x + 1$ $y = -x + 2$

Step 6: Find the Intersection of the Normal and the Curve

To find where the normal intersects the curve again, we substitute $y = -x + 2$ into the equation of the curve: $x^2 + 2x(-x + 2) - 3(-x + 2)^2 = 0$ $x^2 - 2x^2 + 4x - 3(x^2 - 4x + 4) = 0$ $x^2 - 2x^2 + 4x - 3x^2 + 12x - 12 = 0$ $-4x^2 + 16x - 12 = 0$ Divide by -4: $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ So, $x = 1$ or $x = 3$. We already know the normal intersects the curve at $x = 1$. We need to find the other intersection point at $x = 3$.

Step 7: Find the y-coordinate of the Intersection Point

Substitute $x = 3$ into the equation of the normal $y = -x + 2$: $y = -3 + 2 = -1$ The other intersection point is (3, -1).

Step 8: Determine the Quadrant

The point (3, -1) has a positive x-coordinate and a negative y-coordinate, which means it lies in the fourth quadrant.

Common Mistakes & Tips

• Remember to use the product rule and chain rule correctly when performing implicit differentiation.
• Double-check your calculations, especially when substituting and simplifying.
• When finding the intersection points, make sure to consider all possible solutions.

Summary

We found the equation of the normal to the curve $x^2 + 2xy - 3y^2 = 0$ at the point (1, 1) to be $y = -x + 2$. We then found the other intersection point of the normal and the curve to be (3, -1), which lies in the fourth quadrant. Therefore, the normal meets the curve again in the fourth quadrant.

Final Answer

The final answer is \boxed{B}, which corresponds to option (B).