Point P lie on hyperbola `2xy=1`. A triangle is contructed by P, S and S' (where S and S' are foci). The locus of ex-centre opposite S (S and P lie in first quandrant) is `(x+py)^(2)=(sqrt(2)-1)^(2)(x-y)^(2)+q`, then the value of `p+q` is

Answer

To solve the problem, we need to find the locus of the ex-center opposite to the focus S of the hyperbola defined by the equation \(2xy = 1\).

### Step-by-Step Solution:

1. **Identify the Hyperbola and Foci**:
The given hyperbola is \(2xy = 1\). We can rewrite it in standard form:
\[ xy = \frac{1}{2} \]
The foci of this hyperbola are located at \(S(1, 1)\) and \(S'(-1, -1)\).

2. **Parameterize the Point P**:
Let \(P\) be a point on the hyperbola. We can express the coordinates of point \(P\) in terms of a parameter \(t\):
\[ P\left(\frac{t}{\sqrt{2}}, \frac{1}{\sqrt{2}}t\right) \]

3. **Calculate Distances**:
We need to find the distances \(a\), \(b\), and \(c\) for the triangle formed by points \(P\), \(S\), and \(S'\):
- Distance \(a\) (from \(P\) to \(S\)):
\[ a = \sqrt{\left(\frac{t}{\sqrt{2}} - 1\right)^2 + \left(\frac{1}{\sqrt{2}}t - 1\right)^2} \]
- Distance \(b\) (from \(S\) to \(S'\)):
\[ b = 2\sqrt{2} \]
- Distance \(c\) (from \(P\) to \(S'\)):
\[ c = \sqrt{\left(\frac{t}{\sqrt{2}} + 1\right)^2 + \left(\frac{1}{\sqrt{2}}t + 1\right)^2} \]

4. **Find the Ex-center Coordinates**:
The coordinates of the ex-center \(R(h, k)\) opposite to \(S\) can be calculated using the formula:
\[ h = \frac{-a + b + c}{2} \]
\[ k = \frac{-a + b - c}{2} \]

5. **Substituting Values**:
Substitute the expressions for \(a\), \(b\), and \(c\) into the equations for \(h\) and \(k\). After simplification, we will obtain expressions for \(h\) and \(k\) in terms of \(t\).

6. **Find the Locus**:
The locus of the ex-center can be expressed in the form:
\[ (x + py)^2 = ( \sqrt{2} - 1)^2 (x - y)^2 + q \]
By comparing the coefficients, we can find the values of \(p\) and \(q\).

7. **Final Calculation**:
After determining \(p\) and \(q\), we compute \(p + q\).