To the parabola (x+2y)^2+2x-y-3=0 parallel to the line 4x+3y=2.

x^2 + 4xy + 4y^2 + 2x - y - 3 = 0

Take the derivative and solve for dy/dx

2x * dx + 4x * dy + 4y * dx + 8y * dy + 2 * dx - dy = 0
dx * (2x + 4y + 2) + dy * (4x + 8y - 1) = 0
dy * (4x + 8y - 1) = -dx * (2x + 4y + 2)
dy/dx = -(2x + 4y + 2) / (4x + 8y - 1)

Set that to equal to slope of the line (4x + 3y) = 2 and solve for y in terms of x

dy/dx = -4/3

-4/3 = -(2x + 4y + 2) / (4x + 8y - 1)
4 * (4x + 8y - 1) = 3 * (2x + 4y + 2)
2 * (4x + 8y - 1) = 3 * (x + 2y + 1)
8x + 16y - 2 = 3x + 6y + 3
16y - 6y = 3 + 2 + 3x - 8x
10y = 5 - 5x
2y = 1 - x
y = (1 - x) / 2

Now, plug that value for y into the original equation and solve for x

(x + 2 * (1 - x) / 2)^2 + 2x - (1 - x) / 2 - 3 = 0
(x + 1 - x)^2 + 2x + (x - 1) / 2 - 3 = 0
1^2 + 2x + (x/2) - (1/2) - 3 = 0
(5/2) * x = 3 + (1/2) - 1
(5/2) * x = 2 + (1/2)
(5/2) * x = 5/2
x = 1

Solve for y

(1 + 2y)^2 + 2 - y - 3 = 0
1 + 4y + 4y^2 - 1 - y = 0
4y^2 + 3y = 0
y * (4y + 3) = 0
y = 0
y = -3/4

So there are 2 times when the parabola is parallel to the line 4x + 3y = 2

y = (-4/3) * x + b
-3/4 = (-4/3) * 1 + b
-3/4 = -4/3 + b
4/3 - 3/4 = b
7/12 = b