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

Line 4x + 3y = 2 has slope = -4/3

So we need to find point on parabola where dy/dx = -4/3

We find dy/dx using implicit differentiation:

(x + 2y)² + 2x - y - 3 = 0

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

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

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Now we let dy/dx = -4/3
-2(x+2y+1) / (4x+8y-1) = -4/3
2(x+2y+1) / (4x+8y-1) = 4/3
4(4x+8y-1) = 6(x+2y+1)
16x + 32y - 4 = 6x + 12y + 6
10x = 10 - 20y
x = 1 - 2y

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Now we substitute x with (1-2y) in equation of parabola, and solve for y:

(x + 2y)² + 2x - y - 3 = 0
(1-2y + 2y)² + 2(1-2y) - y - 3 = 0
(1)² + 2 - 4y - y - 3 = 0
-5y = 0
y = 0

x = 1-2y = 1

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Line tangent to parabola with slope = -4/3 passes through point (1, 0)

y = -4/3 (x - 1)
y = -4/3 x + 4/3

4x + 3y = 4

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