If `A O Ba n dC O D` are two straight lines which bisect one another at right angles, show that the locus of a points `P` which moves so that `P AxP B=P CxP D` is a hyperbola. Find its eccentricity.

Explanation

To solve the problem, we need to show that the locus of the point \( P \) such that \( PA \times PB = PC \times PD \) is a hyperbola, and then find its eccentricity.

### Step-by-Step Solution:

1. **Define the Points:**
Let the coordinates of the points be:
- \( A(-a, 0) \)
- \( B(a, 0) \)
- \( C(0, b) \)
- \( D(0, -b) \)
- Let \( P \) have coordinates \( (x, y) \).

2. **Calculate Distances:**
We need to calculate the distances \( PA, PB, PC, \) and \( PD \):
- \( PA = \sqrt{(x + a)^2 + y^2} \)
- \( PB = \sqrt{(x - a)^2 + y^2} \)
- \( PC = \sqrt{x^2 + (y - b)^2} \)
- \( PD = \sqrt{x^2 + (y + b)^2} \)

3. **Set Up the Equation:**
According to the problem, we have:
\[ PA \times PB = PC \times PD \]
Substituting the distances, we get:
\[ \sqrt{(x + a)^2 + y^2} \times \sqrt{(x - a)^2 + y^2} = \sqrt{x^2 + (y - b)^2} \times \sqrt{x^2 + (y + b)^2} \]

4. **Square Both Sides:**
Squaring both sides to eliminate the square roots gives:
\[ ((x + a)^2 + y^2)((x - a)^2 + y^2) = (x^2 + (y - b)^2)(x^2 + (y + b)^2) \]

5. **Expand Both Sides:**
Expanding the left-hand side:
\[ ((x^2 + 2ax + a^2 + y^2)(x^2 - 2ax + a^2 + y^2) = (x^2 + y^2 + a^2)^2 - (2ax)^2 \]
This simplifies to:
\[ (x^2 + y^2 + a^2)^2 - 4a^2x^2 \]

Expanding the right-hand side:
\[ (x^2 + y^2 - b^2)^2 - (2by)^2 \]
This simplifies to:
\[ (x^2 + y^2 + b^2)^2 - 4b^2y^2 \]

6. **Set the Equation:**
Now we have:
\[ (x^2 + y^2 + a^2)^2 - 4a^2x^2 = (x^2 + y^2 + b^2)^2 - 4b^2y^2 \]

7. **Rearranging Terms:**
Rearranging gives us a standard form of a hyperbola:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

8. **Identify the Hyperbola:**
This is the equation of a hyperbola centered at the origin.

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

### Conclusion:
Thus, we have shown that the locus of the point \( P \) is a hyperbola, and its eccentricity is \( \sqrt{2} \).