Locus with respect to two intersecting lines

Question

Find locus of a point moving between two lines $ x= \pm y\, \tan \alpha $ so that ratio of distances to them is a constant $= e$

Answer 1

The distance between a point $(u,v)$ and a line $l:ax+by-c=0,$ in the sense of the length of the perpendicular from $(u,v)$ to $l,$ is given by $$d=\frac{|au+bv-c|}{\sqrt{a^2+b^2}}. \tag{1}$$

In your case the two lines have $a=1,b=\pm \tan \alpha.$ So on putting $d_1=d_2 \cdot e,$ the denominators in (1) cancel, and we get $$|x+\tan\alpha y|=e \cdot |x-\tan \alpha y| \tag{2}$$ (or the $+,-$ here are switched).

Now we consider signs for the sides of (2), and see the locus is a line.