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

find the tangent lines as directed.

THANK YOU

eqn of line is

4x + 3y = 2

=> 3y = -4x + 2

=> y = -(4/3)x + (2/3)

so slope of tangent line = - 4/3 (since tangent and given line are parallel)

eqn of parabola

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

differentiating implicitly

2(x + 2y)(1 + 2y ' ) + 2 - y ' = 0

=> 2x + 4y + 4xy ' + 8y y ' + 2 - y ' = 0

=> 2x + 4y + y ' (4x + 8y - 1) + 2 = 0

==> y ' = - (2 + 2x + 4y)/(4x + 8y - 1)

Therefore - (2 + 2x + 4y)/(4x + 8y - 1) = -4/3

6 + 6x + 12y = 16x + 32y - 4

==> 20y = 10 - 10x

y = (1/2)(1 - x)

substitute in the eqn of parabola

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

=> 1+ 2x - 3 - (1/2)(1 - x) = 0

=> 2 + 4x - 6 - 1 + x = 0

=> 5x = 5

x = 1

y = 1/2(1 - x) = 0

point of tangency = (1, 0)

slope = -4/3

eqn of tangent

y - 0 = -4/3(x - 1)

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

3y + 4x = 4 ---------------------------eqn of tangent

You need to find dy/dx using implicit differentiation. I use y' for dy/dx.
2(x+2y)(1+2y' ) + 2 - y'=0.
From the straight line m=-4/3 so y'=-4/3 giving 2(x+2y)(1- 8/3) +2 + 4/3=0 and this cleans up tp
x+2y=1 . Sub into the big equation 1 + 2x-y-3=0 or 2x-y=2
Solve x+2y=1 and 2x-y=2 simultaneously to give x=1, y=0
so the tangent line has equation y=(-4/3)(x-1) or 4x+3y=4.

To differentiate (x+2y)^2 with respect to x , fist sub u=x+2y and let z=u^2.

You want dz/dx= (dz/du)(du/dx) by chain rule
= 2udu/dx
du/dx= 1+dy/dx and you should be able to see the rest.

(x+2y)^2 +2x -y -3 =0 ..................(i)
differentiate
2(x+2y){1+2dy/dx} +2 -dy/dx =0
(2x+4y){1+2*dy/dx} +2 - dy/dx =0
dy/dx{4x+8y-1} +(2x+4y+2) =0
dy/dx = -{2x+4y +2}/(4x+8y-1)............(ii)