The given equation of hyperbola is, H:16x2−9y2=1
⇒e1=1+4232=1616+9
⇒e1=45
Foci of this hyperbola is given by, F(±ae,0).
⇒F(±5,0)
It is given that, e1e2=1
⇒e2=54
Also, ellipse is passing through (±5,0)
⇒a252+b20=1
⇒a225=1
⇒a2=25
⇒a=5
Now, e2=1−a2b2
⇒54=1−25b2
⇒2516=1−25b2
⇒25b2=259
⇒b2=9
∴a=5 and b=3
E:25x2+9y2=1

Putting, y=2
⇒25x2+94=1
⇒25x2=95
⇒x2=9125
⇒x=±355
So, end points of chord are (±355,2)
Thus, length of chord PQ- is given by,
LPQ=2×355
⇒LPQ=3105