Question 1154857
<pre>
We want the equations of the two green lines below

{{{drawing(400,250,-4,4,-2,3,

graph(400,250,-4,4,-2,3,(1/2)(1 - sqrt(5 - 4x))),
green(line(-9,11,7,-5),line(-10,-11,8,7)),circle(1,0,.1),circle(1,1,.1),
graph(400,250,-4,4,-2,3,(1/2)(1 + sqrt(5 - 4x))) )}}}

The y-coordinates of the two points at x=1 are found by setting x=1 and
solving for y:

{{{matrix(6,3,
x + y^2-y,""="",1,
1 + y^2-y,""="",1,
y^2-y,""="",0,
y(y-1),""="",0,
y=0,"",y-1=0,
"","",y=1)}}}

So the two points of tangency are the points (1,0) and (1,1)

We take the derivative implicitly (i.e., without solving for y),
to find the slope m of each of the green tangent lines:

{{{matrix(5,3,
x+y^2-y,""="",1, 
1 +2y*expr(dy/dx)-dy/dx,""="",0,
2y*expr(dy/dx)-dy/dx,""="",-1,
expr(dy/dx)(2y-1),""="",-1,
dy/dx,""="",-1/(2y-1))}}}

So the slope m of the tangent line at (1,0) is 
found by setting y equal to the y-coordinate of
(1,0), which is 0, in the equation for the 
derivative above:

{{{matrix(4,3,
dy/dx,""="",-1/(2y-1),
dy/dx,""="",-1/(2(0)-1),
dy/dx,""="",1,
m,""="",1)}}}

So we are looking for the equation of the green line through 
(1,0) with slope m = 1

{{{matrix(3,3,
y-y[1],""="",m(x-x[1]),
y-0,""="",1(x-1),
y,""="",x-1)}}}

That's the equation of the tangent line at (1,0).

Similarly the slope m of the green tangent line at (1,1) is 
found by setting y equal to the y-coordinate of
(1,1), which is 1, in the equation for the 
derivative above:

{{{matrix(4,3,
dy/dx,""="",-1/(2y-1),
dy/dx,""="",-1/(2(1)-1),
dy/dx,""="",-1,
m,""="",-1)}}}

So we are looking for the equation of the green line through 
(1,1) with slope m = -1

{{{matrix(4,3,
y-y[1],""="",m(x-x[1]),
y-1,""="",-1(x-1),
y-1,""="",-x+1,
y,""="",-x+2
)}}}

That's the equation of the green tangent line at (1,1).

Edwin</pre>