The equation of tangent to the hyperbola
xy=16 at the point P(2) in the first quadrant is:
Given, xy=16,
P(2)⇒t=2
⇒c2=16⇒c=±4
As it lies in first quadrant.
So, we take
⇒c=4
Equation of tangent on hyperbola
xy=c2 at the point P(t) is
x+yt2=2ct
Putting the values of t we get
⇒x+y×22=2×4×2
⇒x+4y=16
Hence, the correct answer is Option (b).