Let r1 is inclined at an angle θ with the axis then coordinates of point are P(r1cosθ,r1sinθ)
Then r2 is inclined at 90+θ with the axis, then coordinates of Q are (r2sinθ,−r2cosθ)
Now P lies on the parabola
(r1sinθ)2=4ar1cosθr1sin2θ=4acosθ......(i)
Q also lies on the parabola
(−r2cosθ)2=4ar2sinθr2cos2θ=4asinθ.......(ii)
Dividing (i) by (ii)
tan3θ=r2r1tanθ=(r2r1)13
cosθ=1√1+tan2θ⇒cosθ=r131√r231+r232sinθ=tanθcosθ⇒sinθ=r132√r231+r232
Substituting sinθ and cosθ in (i)
r1⎛⎜ ⎜ ⎜ ⎜⎝r132√r231+r232⎞⎟ ⎟ ⎟ ⎟⎠2=4ar131√r231+r232r1r232r231+r232=4ar131√r231+r232r231r232=4a√r231+r232
Squaring both sides
r431r432=16a2(r231+r232)