In 2 dimensions, we can draw 2 parallel lines that have the same distance from a line.
I wanted to find parallel functions of a function and their distance is $d$ to the function for all inputs and tangents are equal as shown in the picture.
I assume we have $f(x)$ and we try to find parallel functions that named $g(x)$. $g(x)$ must have 2 solutions $g_1(x)$ and $g_2(x)$ as shown in the picture:
Equations to find g(x):
Equation $(1)$ :Parallel condition $f'(x_1)=g'(x_2)$
Equation $(2)$ : $A(x_1,f(x_1))$ and $B(x_2,g(x_2))$ they are in same line. $g(x_2)-f(x_1)=\frac{-1}{f'(x_1)}(x_2-x_1)$
Equation $(3)$ : d is between $A(x_1,f(x_1))$ and $B(x_2,g(x_2))$ $d^2=(x_2-x_1)^2+(g(x_2)-f(x_1))^2$
$(x_2-x_1)^2+(\frac{1}{(f'(x_1))^2}(x_2-x_1)^2=d^2$
$(x_2-x_1)^2+\frac{1}{(f'(x_1))^2}(x_2-x_1)^2=d^2$
$(x_2-x_1)^2==\frac{d^2(f'(x_1))^2}{1+(f'(x_1))^2} $
$x_2-x_1=+\frac{d.f'(x_1)}{\sqrt{1+(f'(x_1))^2}} $
$x_2-x_1=-\frac{d.f'(x_1)}{\sqrt{1+(f'(x_1))^2}} $
if we want to find first solution of $g(x)$
then need to take $x_2=x_1+\frac{df'(x_1)}{\sqrt{1+(f'(x_1))^2}} $ and put in Equation (1)
$f'(x_1)=g'(x_1+\frac{d.(f'(x_1))}{\sqrt{1+(f'(x_1))^2}})$ replace $x_1$ with $x$ means for all values . and I tried to find $g(x)$ $f'(x)=g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$
$f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})')=(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$
$\int f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx= \int (1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})'). g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}}) dx$
$\int f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx= g(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$
$g(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})=f(x)+\int f'(x)(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx$
Am I in the right way to find $g(x)$? Can I find g(x) after integrations? Thanks in advance.