I was looking at this solution to an exercise (which I can provide if necessary):
$$T=\frac{(x',y')}{\sqrt{x'^2+y'^2}}$$ $$N=\frac{(y',-x')}{\sqrt{x'^2+y'^2}}$$ $$(x,y)+(\pm d)N=(x,y)+\frac{(\pm d)(y',-x')}{\sqrt{x'^2+y'^2}}=(x_2,0)$$ $$x+\frac{((\pm)d)y'}{\sqrt{x'^2+y'^2}}=x_2$$ $$y+\frac{((\pm)d)(-x')}{\sqrt{x'^2+y'^2}}=0$$ $$\lvert x_2-x\rvert=2=\left\lvert x+\frac{\pm dy'}{\sqrt{x'^2+y'^2}}-x\right\rvert=\frac{\lvert d\rvert\lvert y'\rvert}{\sqrt{x'^2+y'^2}}$$ Divide the above by $2$ by the expression for $\displaystyle y\implies\frac{\lvert d\rvert\lvert y'\rvert}{\pm dx'}=\frac{2}{y}$ $$\pm\frac{dy}{dx}=\frac{2}{y}\implies\pm ydy=2dx$$ $$\implies\pm\frac{1}{2}(y_f^2-y_i^2)=2(x_f-x_i)$$ $$\underrightarrow{(1,2)}\pm(y^2-4)=4(x-1)\implies\boxed{\begin{array}{c}y^2=4x\\y^2=-4x+8\end{array}}$$
Solution to Exercise 14 (14.9 Apostol) (Image that replaced text)
And I am perplexed as to how they go from the line that starts with "Divide by 2" to the next line. It doesn't appear that they are integrating, yet somehow they manage to turn $x'$ into $x$ and $y'$ into $y$. Also, what about the line after that, I have no idea how they get to that either. Last line is just substituting in a point.
The instruction is not "Divide by $2$." It is "Divide the above $2$ [from the previous line] by the expression for $y$ [from the line before that]." That is, divide $$ 2 = \frac{|d| \, |y'|}{\sqrt{x'^2 + y'^2}} $$ by $$ y = - \frac{(\pm d)(-x')}{\sqrt{x'^2 + y'^2}} \text{,} $$ yielding $$ \frac{2}{y} = \frac{|d| \, |y'|}{-(\pm d)(-x')} = \frac{|d| \, |y'|}{\pm dx'} \text{.} $$
This is one of $\pm \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$, where I have used $t$ as the independent parameter for $x$, $y$, and the vectors $T$ and $N$. We should recognize this as (up to sign) the slope of the tangent line to the curve through the point $(x(t), y(t))$. That is, this is $\pm \frac{\mathrm{d}y}{\mathrm{d}x}$.
Then they proceed by separation of variables and integration.