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 \(P\) is any point on the hyperbola \(x^2 - y^2 = a^2\) and that \(PF_1 \cdot PF_2 = \lambda (OP)^2\), where \(O\) is the origin.

### Step-by-Step Solution:

1. **Identify the Hyperbola and Its Properties**:
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 \(b^2 = a^2\).

2. **Determine the Eccentricity**:
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \]

3. **Express Distances from Point \(P\)**:
Let \(P\) be a point on the hyperbola with coordinates \((x_1, y_1)\). The distances from \(P\) to the foci are:
\[ PF_1 = e(x_1 - a) = \sqrt{2}(x_1 - a) \]
\[ PF_2 = e(x_1 + a) = \sqrt{2}(x_1 + a) \]

4. **Calculate the Product of Distances**:
Now, we calculate the product \(PF_1 \cdot PF_2\):
\[ PF_1 \cdot PF_2 = \left(\sqrt{2}(x_1 - a)\right) \cdot \left(\sqrt{2}(x_1 + a)\right) = 2(x_1^2 - a^2) \]

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

6. **Substitute into the Given Equation**:
We have the equation:
\[ PF_1 \cdot PF_2 = \lambda (OP)^2 \]
Substituting the expressions we found:
\[ 2(x_1^2 - a^2) = \lambda (x_1^2 + y_1^2) \]

7. **Use the Hyperbola Equation**:
From the hyperbola equation \(x_1^2 - y_1^2 = a^2\), we can express \(y_1^2\) in terms of \(x_1^2\):
\[ y_1^2 = x_1^2 - a^2 \]

8. **Substitute \(y_1^2\) into the Equation**:
Substitute \(y_1^2\) into the equation:
\[ 2(x_1^2 - a^2) = \lambda \left(x_1^2 + (x_1^2 - a^2)\right) \]
Simplifying gives:
\[ 2(x_1^2 - a^2) = \lambda (2x_1^2 - a^2) \]

9. **Solve for \(\lambda\)**:
Rearranging the equation:
\[ 2(x_1^2 - a^2) = \lambda (2x_1^2 - a^2) \]
Dividing both sides by \(2x_1^2 - a^2\) (assuming \(x_1^2 \neq \frac{a^2}{2}\)):
\[ \lambda = 1 \]

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