The fouce of the rectangular hyperbola `(x-h) (y-k) =p^(2) ` is

Options

A. `(h -p, k -p)`
B. `(h -p, k +p)`
C. `(h+ p, k -p)`
D. none of these

Answer

To find the foci of the rectangular hyperbola given by the equation \((x-h)(y-k) = p^2\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation of the rectangular hyperbola:
\[ (x-h)(y-k) = p^2 \]
Let \(X = x - h\) and \(Y = y - k\). Then the equation becomes:
\[ XY = p^2 \]

### Step 2: Identify the foci of the standard hyperbola
For the standard form of a rectangular hyperbola \(XY = c^2\), the foci are located at \((c, c)\) and \((-c, -c)\). Here, \(c\) corresponds to the value of \(p\), so the foci are:
\[ (p, p) \quad \text{and} \quad (-p, -p) \]

### Step 3: Transform back to original coordinates
Now, we need to convert these foci back to the original coordinates \((x, y)\). Using the transformations \(X = x - h\) and \(Y = y - k\):
1. For the focus \((p, p)\):
\[ x = p + h \quad \text{and} \quad y = p + k \]
Thus, one focus is \((p+h, p+k)\).

2. For the focus \((-p, -p)\):
\[ x = -p + h \quad \text{and} \quad y = -p + k \]
Thus, the other focus is \((-p+h, -p+k)\).

### Final Result
The foci of the rectangular hyperbola \((x-h)(y-k) = p^2\) are:
\[ (p+h, p+k) \quad \text{and} \quad (-p+h, -p+k) \]