Explanation:

To find the point(s) where the normal to the curve intersects the curve again, we need to find the slope of the tangent line to the curve at the given point (1,1) and then find the equation of the normal line passing through this point.

1) Finding the slope of the tangent line:
To find the slope of the tangent line at a given point, we take the derivative of the curve equation with respect to x and evaluate it at that point.

The given curve equation is: x^2 + 2xy - 3y^2 = 0

Taking the derivative of both sides with respect to x gives:
2x + 2y(dx/dx) + 2x(dy/dx) - 6y(dy/dx) = 0

Simplifying, we get:
2x + 2y + 2x(dy/dx) - 6y(dy/dx) = 0

At the point (1,1), we substitute x=1 and y=1 to get:
2(1) + 2(1) + 2(1)(dy/dx) - 6(1)(dy/dx) = 0
2 + 2 + 2(dy/dx) - 6(dy/dx) = 0
4 - 4(dy/dx) = 0
(dy/dx) = 1

Therefore, the slope of the tangent line at the point (1,1) is 1.

2) Finding the equation of the normal line:
The normal to the curve is perpendicular to the tangent line at the point of contact. Since the slope of the tangent line is 1, the slope of the normal line will be -1 (negative reciprocal).

Using the point-slope form of a line, we can write the equation of the normal line passing through (1,1) as:
y - 1 = -1(x - 1)
y - 1 = -x + 1
y = -x + 2

3) Finding the point(s) of intersection:
To find the point(s) of intersection, we substitute the equation of the normal line into the curve equation and solve for x and y.

Substituting y = -x + 2 into the curve equation x^2 + 2xy - 3y^2 = 0, we get:
x^2 + 2x(-x + 2) - 3(-x + 2)^2 = 0
x^2 - 2x^2 + 4x + 3x^2 - 12x + 12 = 0
2x^2 - 8x + 12 = 0
x^2 - 4x + 6 = 0

This quadratic equation has no real roots, indicating that the normal line does not intersect the curve again in the first quadrant.

4) Conclusion:
Since the normal line does not intersect the curve again in the first quadrant, it must intersect the curve again in one of the other three quadrants. Therefore, the correct answer is option D) meet the curve again in the fourth quadrant.