If the tangent drawn to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` at any point P meets the coordinate axes at the points A and B respectively. If the rectangle OACB (O being the origin) is completed, where C lies on `(x^2)/(a^2)-(y^2)/(b^2)=lambda` . Then, the value of `lambda` is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the equation of the tangent to the hyperbola
The equation of the hyperbola is given as:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

For a point \( P(a \sec \theta, b \tan \theta) \) on the hyperbola, the equation of the tangent at point \( P \) is given by:

\[ \frac{x \cdot \sec \theta}{a} - \frac{y \cdot \tan \theta}{b} = 1 \]

### Step 2: Find the points where the tangent meets the axes
To find the points where the tangent meets the coordinate axes, we will set \( y = 0 \) to find point \( A \) on the x-axis and \( x = 0 \) to find point \( B \) on the y-axis.

1. **Finding point A (where \( y = 0 \))**:
\[ \frac{x \cdot \sec \theta}{a} - 0 = 1 \implies x = a \sec \theta \]
Thus, point \( A \) is \( \left(a \sec \theta, 0\right) \).

2. **Finding point B (where \( x = 0 \))**:
\[ 0 - \frac{y \cdot \tan \theta}{b} = 1 \implies y = -b \tan \theta \]
Thus, point \( B \) is \( \left(0, -b \tan \theta\right) \).

### Step 3: Form the rectangle OACB
The rectangle OACB is formed with points \( O(0,0) \), \( A(a \sec \theta, 0) \), \( B(0, -b \tan \theta) \), and \( C(a \sec \theta, -b \tan \theta) \).

### Step 4: Use the coordinates of point C to find the value of \( \lambda \)
Point \( C \) lies on the hyperbola defined by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = \lambda \]

Substituting the coordinates of point \( C(a \sec \theta, -b \tan \theta) \):

\[ \frac{(a \sec \theta)^2}{a^2} - \frac{(-b \tan \theta)^2}{b^2} = \lambda \]

This simplifies to:

\[ \sec^2 \theta - \tan^2 \theta = \lambda \]

Using the trigonometric identity \( \sec^2 \theta - \tan^2 \theta = 1 \):

\[ \lambda = 1 \]

### Final Result
Thus, the value of \( \lambda \) is:

\[ \boxed{1} \]