Equation of the rectangular hyperbola whose focus is `(1,-1)` and the corresponding directrix is `x-y+1=0`.
a.`x^(2)-y^(2)=1` b.`xy=1` c.`2xy-4x+4y+1=0` d.`2xy+4x-4y-1=0`

Options

A. `x^(2)-y^(2)=1`
B. `xy=1`
C. `2xy-4x+4y+1=0`
D. `2xy+4x-4y-1=0`

Explanation

To find the equation of the rectangular hyperbola whose focus is at \( (1, -1) \) and whose directrix is given by the line \( x - y + 1 = 0 \), we can follow these steps:

### Step 1: Identify the focus and directrix
The focus \( S \) is given as \( (1, -1) \) and the directrix \( D \) is given by the equation \( x - y + 1 = 0 \).

### Step 2: Use the definition of a hyperbola
For a hyperbola, the distance from any point \( P(x, y) \) on the hyperbola to the focus \( S \) is \( e \) times the distance from \( P \) to the directrix \( D \), where \( e \) is the eccentricity. For a rectangular hyperbola, \( e = \sqrt{2} \).

### Step 3: Write the distance formulas
The distance \( PS \) from point \( P(x, y) \) to the focus \( S(1, -1) \) is given by:
\[ PS = \sqrt{(x - 1)^2 + (y + 1)^2} \]

The distance \( PM \) from point \( P(x, y) \) to the directrix can be calculated using the formula for the distance from a point to a line. The distance from point \( P(x, y) \) to the line \( ax + by + c = 0 \) is given by:
\[ PM = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
For the line \( x - y + 1 = 0 \), we have \( a = 1, b = -1, c = 1 \). Thus,
\[ PM = \frac{|1 \cdot x - 1 \cdot y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 1|}{\sqrt{2}} \]

### Step 4: Set up the equation using the definition of hyperbola
According to the definition of hyperbola:
\[ PS = e \cdot PM \]
Substituting the distances we found:
\[ \sqrt{(x - 1)^2 + (y + 1)^2} = \sqrt{2} \cdot \frac{|x - y + 1|}{\sqrt{2}} \]
This simplifies to:
\[ \sqrt{(x - 1)^2 + (y + 1)^2} = |x - y + 1| \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - 1)^2 + (y + 1)^2 = (x - y + 1)^2 \]

### Step 6: Expand both sides
Expanding the left side:
\[ (x^2 - 2x + 1) + (y^2 + 2y + 1) = x^2 + y^2 - 2x + 2y + 2 \]
The left side simplifies to:
\[ x^2 + y^2 - 2x + 2y + 2 \]

Expanding the right side:
\[ (x - y + 1)^2 = x^2 - 2xy + y^2 + 2x - 2y + 1 \]

### Step 7: Set the expanded equations equal
Setting both expansions equal:
\[ x^2 + y^2 - 2x + 2y + 2 = x^2 - 2xy + y^2 + 2x - 2y + 1 \]

### Step 8: Simplify the equation
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x + 2y + 2 = -2xy + 2x - 2y + 1 \]
Rearranging gives:
\[ 2xy - 4x + 4y + 1 = 0 \]

### Final Equation
Thus, the equation of the rectangular hyperbola is:
\[ 2xy - 4x + 4y + 1 = 0 \]

### Conclusion
The correct answer is option **c. \( 2xy - 4x + 4y + 1 = 0 \)**.

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