From standard Newtonian form for focal conics $ p/r = ( 1- \epsilon \cos \theta), $ I obtained by differentiating with respect to arc:
$$ \dfrac{FN}{p} = \dfrac{\cos \phi}{\sin \theta}. $$
where $ FN $= perpendicular length dropped from focus F on tangent at P and, $p$ is the semi-Latus Rectum. Please suggest a geometric construction for its proof.
EDIT 1:
I assumed drawing new construction lines would make one see a geometrical relationship.
I am unable to immediately think out a line of common shared length $$ FN \sin \theta = p \cos \phi $$
or a line of length $$ FN /\cos \phi = p /sin\theta $$ or any other.
EDIT2:
$ (p \cos \phi) $ does not change even if we consider the other focus.It may relate to reflection angle between foci at NPF to the tangent.
EDIT3:
The following construction I have completed in continuation with Lee David Chung Lin's setting as an aid. Take the variable point $T$ as the generating circle diameter. Draw a circle with $F$ as center and $p$ as radius. Angle $\theta $ is both an alternate angle between two parallel lines as well as it is contained in the alternate segment making $ FST, FBR $ triangles congruent. $TS$ is perpendicular to ellipse radius vector $FP$ and tangent to circle of radius $p$ at $S$.
$$ FS= p= FT \,\sin \theta $$
$$ FN= FT \,\cos \phi $$
Dividing
$$ \dfrac{FN}{p} = \dfrac{\cos \phi}{\sin \theta} $$
