A rod of length $l$ slides on two mutually perpendicular axes. At the ends of the rod, two lines making angles $60^\circ$ and $30^\circ$ with the rod are drawn then prove that the locus of point of intersection of the lines is independent of the length of the rod. Also, find the position of maximum distance of the locus from the origin.
Assuming that the rod is sliding on x-axis and y-axis. When it is on x-axis, if one end is at origin O, the other end is at A and one line making $60^\circ$ with the rod at O and other line making $30^\circ$ at A then that means the point of intersection is on the line passing through origin and making $60^\circ$ with x-axis. Thus, locus is $y=\sqrt3x$.
Similarly, if line through O is making $30^\circ$ with the rod and the one through A is making $60^\circ$ with the rod then locus will be $y=\frac1{\sqrt3}x$
How to find the position of maximum distance of the locus from the origin?
In the solution given, they have said angle BAO=$60^\circ$ (B is the end point of the rod when it is moving on y-axis)
And angle OBA=$30^\circ$ or OP=BA=$l$.
I do not understand this.
A similar question exists on this website. But there, the rod is sliding between two axes. In my question, it's on the two axes. Also, in my question, the locus is independent of the length of the rod. But in the linked question, locus is dependent on the length of the rod.