Area of the triangle formed by any arbitrary tangents of the hyperbola `xy = 4`, with the co-ordinate axes is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To find the area of the triangle formed by any arbitrary tangents of the hyperbola \( xy = 4 \) with the coordinate axes, we can follow these steps:

### Step 1: Identify the point on the hyperbola
Let the point \( P(x_1, y_1) \) be on the hyperbola. Since \( P \) lies on the hyperbola, it satisfies the equation:
\[ x_1 y_1 = 4 \]

### Step 2: Write the equation of the tangent line
The equation of the tangent to the hyperbola \( xy = 4 \) at the point \( P(x_1, y_1) \) can be derived using the formula for the tangent line. The equation of the tangent line at point \( P \) is given by:
\[ y_1 x + x_1 y = 4 \]

### Step 3: Find the x-intercept (OA)
To find the x-intercept of the tangent line (where \( y = 0 \)), substitute \( y = 0 \) into the tangent equation:
\[ y_1 x + x_1 (0) = 4 \implies y_1 x = 4 \implies x = \frac{4}{y_1} \]
Thus, the x-intercept \( OA \) is:
\[ OA = \frac{4}{y_1} \]

### Step 4: Find the y-intercept (OB)
To find the y-intercept of the tangent line (where \( x = 0 \)), substitute \( x = 0 \) into the tangent equation:
\[ y_1 (0) + x_1 y = 4 \implies x_1 y = 4 \implies y = \frac{4}{x_1} \]
Thus, the y-intercept \( OB \) is:
\[ OB = \frac{4}{x_1} \]

### Step 5: Calculate the area of the triangle
The area \( A \) of the triangle formed by the x-intercept, y-intercept, and the origin is given by:
\[ A = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \frac{4}{y_1} \times \frac{4}{x_1} \]
Substituting \( x_1 y_1 = 4 \) into the area formula:
\[ A = \frac{1}{2} \times \frac{16}{x_1 y_1} = \frac{1}{2} \times \frac{16}{4} = \frac{1}{2} \times 4 = 2 \]

### Conclusion
The area of the triangle formed by any arbitrary tangents of the hyperbola \( xy = 4 \) with the coordinate axes is:
\[ \text{Area} = 8 \text{ square units} \]