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Question

The equation of tangent to the hyperbola
xy=16 at the point P(2) in the first quadrant is:

A
2x+y=16
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B
x+4y=16
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C
x+2y=16
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D
2x+y=8
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Solution

The correct option is B x+4y=16

Given, xy=16,
P(2)t=2
c2=16c=±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).


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