The normal to the curve $$x^2 + 2xy - 3y^2 = 0$$, at (1, 1)

The given curve is $$ x^2 + 2xy - 3y^2 = 0 $$. This equation can be factorized as follows:

Recognize that $$ x^2 + 2xy - 3y^2 $$ is a quadratic form. Factor it by splitting the middle term or by inspection:

$$ x^2 + 2xy - 3y^2 = (x + 3y)(x - y) = 0 $$

Verification: Expand $$ (x + 3y)(x - y) = x \cdot x + x \cdot (-y) + 3y \cdot x + 3y \cdot (-y) = x^2 - xy + 3xy - 3y^2 = x^2 + 2xy - 3y^2 $$, which matches the original equation. Thus, the curve consists of two straight lines: $$ x + 3y = 0 $$ and $$ x - y = 0 $$.

The point (1, 1) is given. Check which line it lies on:

For $$ x - y = 0 $$: $$ 1 - 1 = 0 $$, so it satisfies. For $$ x + 3y = 0 $$: $$ 1 + 3 \cdot 1 = 4 \neq 0 $$, so it does not satisfy. Therefore, (1, 1) lies on the line $$ x - y = 0 $$.

To find the normal to the curve at (1, 1), first find the tangent to the curve at this point. Since the curve is degenerate (two lines) and the point is on $$ x - y = 0 $$, the tangent is the line itself. The slope of $$ x - y = 0 $$ (or $$ y = x $$) is 1.

The normal is perpendicular to the tangent, so its slope is the negative reciprocal of 1, which is -1.

Using the point-slope form, the equation of the normal passing through (1, 1) is:

$$ y - 1 = (-1)(x - 1) $$

Simplify:

$$ y - 1 = -x + 1 $$

$$ y = -x + 1 + 1 $$

$$ y = -x + 2 $$

Now, find where this normal $$ y = -x + 2 $$ intersects the curve $$ (x + 3y)(x - y) = 0 $$ again, besides (1, 1). Substitute $$ y = -x + 2 $$ into each factor of the curve equation.

First, substitute into $$ x - y = 0 $$:

$$ x - (-x + 2) = 0 $$

$$ x + x - 2 = 0 $$

$$ 2x - 2 = 0 $$

$$ 2x = 2 $$

$$ x = 1 $$

Then $$ y = -1 + 2 = 1 $$, giving the point (1, 1).

Next, substitute into $$ x + 3y = 0 $$:

$$ x + 3(-x + 2) = 0 $$

$$ x - 3x + 6 = 0 $$

$$ -2x + 6 = 0 $$

$$ -2x = -6 $$

$$ x = 3 $$

Then $$ y = -3 + 2 = -1 $$, giving the point (3, -1).

Verify that (3, -1) lies on the curve:

$$ x^2 + 2xy - 3y^2 = 3^2 + 2 \cdot 3 \cdot (-1) - 3 \cdot (-1)^2 = 9 - 6 - 3 \cdot 1 = 9 - 6 - 3 = 0 $$

It satisfies the equation. The point (3, -1) has $$ x > 0 $$ and $$ y < 0 $$, so it is in the fourth quadrant.

Therefore, the normal meets the curve again at (3, -1) in the fourth quadrant.

Now, consider the options:

A. Meets the curve again in the fourth quadrant

B. Does not meet the curve again

C. Meets the curve again in the second quadrant

D. Meets the curve again in the third quadrant

Hence, the correct answer is Option A.