Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and co-ordinate axes equals

Options

A. 8
B. 16
C. 32
D. 64

Answer

To find the area of the triangle formed by the tangent to the hyperbola \(xy = 16\) at the point \((16, 1)\) and the coordinate axes, we can follow these steps:

### Step 1: Find the equation of the tangent line to the hyperbola at the given point.
The equation of the hyperbola is given by \(xy = 16\). To find the slope of the tangent line at the point \((16, 1)\), we first differentiate the equation implicitly.

\[ \frac{dy}{dx} = -\frac{y}{x} \]

Substituting the point \((16, 1)\) into the derivative:

\[ \frac{dy}{dx} = -\frac{1}{16} \]

### Step 2: Write the equation of the tangent line.
Using the point-slope form of a line, the equation of the tangent line at the point \((16, 1)\) is:

\[ y - 1 = -\frac{1}{16}(x - 16) \]

Expanding this gives:

\[ y - 1 = -\frac{1}{16}x + 1 \]

Rearranging it, we have:

\[ \frac{1}{16}x + y - 2 = 0 \]

Multiplying through by 16 to eliminate the fraction:

\[ x + 16y - 32 = 0 \]

### Step 3: Find the intercepts with the coordinate axes.
To find the x-intercept, set \(y = 0\):

\[ x + 16(0) - 32 = 0 \implies x = 32 \]

To find the y-intercept, set \(x = 0\):

\[ 0 + 16y - 32 = 0 \implies 16y = 32 \implies y = 2 \]

### Step 4: Calculate the area of the triangle formed by the tangent and the axes.
The area \(A\) of a triangle formed by the x-intercept and y-intercept can be calculated using the formula:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is the x-intercept (32) and the height is the y-intercept (2):

\[ A = \frac{1}{2} \times 32 \times 2 = \frac{64}{2} = 32 \]

Thus, the area of the triangle formed by the tangent to the hyperbola at the point \((16, 1)\) and the coordinate axes is:

\[ \boxed{32} \]