Given equation of ellipse is, 9x2+25y2=1
⇒a=3,b=5
We know that, e=1−b2a2
⇒e=1−259=54
Now, foci=(0,±be)
=(0,±4)
Since, eccentricity of hyperbola is given as 815 same to that of ellipse, ∴eH=54×815=23
Let equation of the hyperbola be A2x2−B2y2=−1.
⇒B.eH=4
⇒B=38
⇒A2=B2(eH2−1)=964(49−1)
⇒A2=980
⇒980x2−964y2=−1
Directrix: y=±eHB=±916
⇒PS=e⋅PM
⇒PS=23∣314⋅52−916∣
⇒PS=752−38