Given P find R

The point Q = (r c₂, r s₂, 2 h - e) is on the line through the R and P. Thus $(Q-R) \parallel (R-P)$ gives $(r c_2 - c, r s_2 - s, h -e) \parallel (c-a, s-b, h)$ which gives

$$\frac {r c_2 - c}{c-a} = \frac {r s_2 - s}{s-b} = \frac {h-e}{h}$$ (9)

The first of the above equalities gives (s-b)(r (2 c² - 1) - c) = (c-a)(2 r c s - s) which gives

s(2 r a c - a - r) = b ( 2 r c² - r - c). (10)

Squaring both sides and substituting 1-c² for s² gives (1-c²)(2 a r c - a - r)² = b² (2 r c² - r - c)². This expands to a quartic equation in the variable c:

4 r² (a² + b²) c⁴ - 4 r (a² + a r + b²) c³ +((a² + b²) (1 - 4 r²) + 2 x r + r²) c² + 2 r (2 a² + 2 a r + b²) c + r² b² - (a+r)² = 0. (11)

For each solution of this equation, a corresponding s can be found from ( 10), and h can be found from ( 9) as follows:

$$\alpha \equiv \frac {r c_2 - c}{c-a} = \frac {h-e}{h}$$ (12)

which gives $h = e / (1-\alpha)$. Out of these solutions of the equations, the one that actually solves our geometrical problem will need to be selected.