The correct options are
C always positive
D always same
Let Z and Z′ are the two directrix of this Hyperbola right hand-side and left hand-side respectively.
As we know Directrix of Hyperbola is always parallel to Conjugate axis or minor axis of hyperbola. The distance of directrix of hyperbola from the center is ae , both sides.
So equation of Z is x=ae and equation of Z′ is x=−ae
Now if a perpendicular is drawn from point P to Both Directrix, let it meet Z at M and Z′ at M′.
Then according to definition of Hyperbola F2P=ePM and F1P=ePM′
Now let the coordinates of point P are (x,y)
Then PM=x−ae
and PM′=x−(−ae)=x+ae
We know that F2P=ePM=e(x−ae)
→ F2P=ex−a ......(1)
Similiarly F1P=ePM′=e(x+ae)
→ F1P=ex+a .....(2)
Subtracting eq. (1) from eq. (2) We get, F1P−F2P=ex+a−(ex−a)
→ F1P−F2P=2a
As for a Hyperbola a>0 so the value of F1P−F2P is always Positive and also Constant as 2a is a constant value.
Hence option A and C are correct.