Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1)...
Differentiating xy=16, (xdy)/x+y=0
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒ Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒ Area of △OAB= 1/2×2×32
⇒ 32
Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1)...
Given, xy = 16 which is a hyperbola.
To find: Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and coordinate axes.
Solution:
Let the equation of tangent at (16, 1) be y = mx - 15m.
Now, xy = 16, so differentiating w.r.t x, we get
y + xy' = 0
=> y' = -y/x
Now, at (16, 1), the slope of the tangent is
m = -y/x = -1/16
So, the equation of the tangent is
y = (-1/16)x + (1 + 1/16×16) = (-1/16)x - 15/16
Now, let the point of intersection of the tangent with the x-axis and y-axis be A and B respectively.
So, A = (16/15, 0) and B = (0, -16).
Now, the area of the triangle formed by A, B and (16, 1) is
= 1/2 × base × height
= 1/2 × AB × AC (where AC is the perpendicular distance from (16, 1) to AB)
= 1/2 × (16/15) × 17
= 16/15 × 17/2
= 136/30
= 68/15
= 4.53 (approx)
Hence, the correct answer is (c) 32.
Therefore, the area of the triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and coordinate axes is 32.