If the polars of `(x_1,y_1)" and "(x_2,y_2)` with respect to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` are at the right angels, where `(x_1x_2)b^4+a^4(y_1y_2)=lambda`, then `lambda` the value of is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( \lambda \) given that the polars of the points \( (x_1, y_1) \) and \( (x_2, y_2) \) with respect to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are at right angles. We are also given the equation \( x_1 x_2 b^4 + a^4 y_1 y_2 = \lambda \).

### Step-by-Step Solution:

1. **Write the equations of the polars**:
The polar of the point \( (x_1, y_1) \) with respect to the hyperbola is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Similarly, the polar of the point \( (x_2, y_2) \) is:
\[ \frac{xx_2}{a^2} - \frac{yy_2}{b^2} = 1 \]

2. **Find the slopes of the polars**:
The slope of the first polar can be derived from the equation:
\[ \frac{yy_1}{b^2} = \frac{xx_1}{a^2} - 1 \implies y = \frac{b^2}{a^2} \cdot \frac{x_1}{y_1} x + \frac{b^2}{y_1} \]
Thus, the slope \( m_1 \) of the first polar is:
\[ m_1 = \frac{b^2}{a^2} \cdot \frac{x_1}{y_1} \]
Similarly, for the second polar:
\[ m_2 = \frac{b^2}{a^2} \cdot \frac{x_2}{y_2} \]

3. **Use the condition for perpendicular lines**:
Since the polars are at right angles, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(\frac{b^2}{a^2} \cdot \frac{x_1}{y_1}\right) \cdot \left(\frac{b^2}{a^2} \cdot \frac{x_2}{y_2}\right) = -1 \]
Simplifying this gives:
\[ \frac{b^4 x_1 x_2}{a^4 y_1 y_2} = -1 \]

4. **Rearranging the equation**:
From the above equation, we can rearrange it to find:
\[ b^4 x_1 x_2 + a^4 y_1 y_2 = 0 \]
This implies:
\[ b^4 x_1 x_2 = -a^4 y_1 y_2 \]

5. **Identify \( \lambda \)**:
We are given that:
\[ x_1 x_2 b^4 + a^4 y_1 y_2 = \lambda \]
From our previous result, we know that \( b^4 x_1 x_2 + a^4 y_1 y_2 = 0 \). Therefore:
\[ \lambda = 0 \]

### Final Answer:
Thus, the value of \( \lambda \) is:
\[ \lambda = 0 \]