The correct option is C 8x−2y−75=0
Given, xy=25
P(2)⇒t=2
⇒c2=25⇒c=5
as it lies in first quadrant
Equation of normal to the hyperbola xy=c2 at the point P(t) is xt3−yt−ct4+c=0
Put the values of t and c
⇒x(2)3−y(2)−c(2)4+5=0
⇒8x−2y−16×5+5=0
⇒8x−2y−75=0
Hence, the correct answer is Option a.