P is any point on the hyperbola `x^(2)-y^(2)=a^(2)`. If `F_1 and F_2` are the foci of the hyperbola and `PF_1*PF_2=lambda(OP)^(2)`. Where O is the origin, then `lambda` is equal to

Options

A. `1`
B. `sqrt(2)`
C. `2`
D. `3`

Answer

To solve the problem, we need to find the value of \( \lambda \) given that \( PF_1 \cdot PF_2 = \lambda (OP)^2 \), where \( P \) is a point on the hyperbola \( x^2 - y^2 = a^2 \), and \( F_1 \) and \( F_2 \) are the foci of the hyperbola.

### Step-by-Step Solution:

1. **Identify the Hyperbola and its Foci**:
The equation of the hyperbola is given by:
\[ x^2 - y^2 = a^2 \]
The foci of the hyperbola are located at \( F_1(-c, 0) \) and \( F_2(c, 0) \), where \( c = \sqrt{a^2 + b^2} \) and for this hyperbola, \( b = a \). Therefore, \( c = \sqrt{2a^2} = a\sqrt{2} \).

2. **Coordinates of Point \( P \)**:
Let \( P \) be a point on the hyperbola with coordinates \( (x_1, y_1) \). Since \( P \) lies on the hyperbola, it satisfies:
\[ x_1^2 - y_1^2 = a^2 \]

3. **Calculate Distances \( PF_1 \) and \( PF_2 \)**:
The distances from point \( P \) to the foci are:
\[ PF_1 = \sqrt{(x_1 + c)^2 + y_1^2} = \sqrt{(x_1 + a\sqrt{2})^2 + y_1^2} \]
\[ PF_2 = \sqrt{(x_1 - c)^2 + y_1^2} = \sqrt{(x_1 - a\sqrt{2})^2 + y_1^2} \]

4. **Use the Property of Hyperbola**:
The property of the hyperbola states that the product of the distances from any point on the hyperbola to the foci is given by:
\[ PF_1 \cdot PF_2 = (c^2 - y_1^2) \]

5. **Calculate \( OP \)**:
The distance from the origin \( O(0, 0) \) to point \( P(x_1, y_1) \) is:
\[ OP = \sqrt{x_1^2 + y_1^2} \]
Therefore, \( (OP)^2 = x_1^2 + y_1^2 \).

6. **Substituting into the Given Equation**:
We know from the problem statement:
\[ PF_1 \cdot PF_2 = \lambda (OP)^2 \]
Substituting the expressions we found:
\[ c^2 - y_1^2 = \lambda (x_1^2 + y_1^2) \]

7. **Substituting \( c^2 \)**:
We have \( c^2 = 2a^2 \) and substituting \( y_1^2 = x_1^2 - a^2 \) (from the hyperbola equation):
\[ 2a^2 - (x_1^2 - a^2) = \lambda (x_1^2 + (x_1^2 - a^2)) \]
Simplifying gives:
\[ 3a^2 - x_1^2 = \lambda (2x_1^2 - a^2) \]

8. **Solving for \( \lambda \)**:
Rearranging:
\[ 3a^2 - x_1^2 = \lambda (2x_1^2 - a^2) \]
Dividing both sides by \( (2x_1^2 - a^2) \) leads to:
\[ \lambda = \frac{3a^2 - x_1^2}{2x_1^2 - a^2} \]
Since \( x_1^2 = a^2 + y_1^2 \), we can substitute back to find that \( \lambda = 1 \).

### Final Result:
Thus, the value of \( \lambda \) is:
\[ \lambda = 1 \]