Find the equation of the hyperbola whose one focus is `(-1, 1)` , eccentricity = 3 and the equation of the corresponding directrix is `x - y + 3 = 0 ` .

Answer

To find the equation of the hyperbola with the given focus, eccentricity, and directrix, we can follow these steps:

### Step 1: Identify the given information
- Focus \( S = (-1, 1) \)
- Eccentricity \( e = 3 \)
- Directrix equation: \( x - y + 3 = 0 \)

### Step 2: Write the general formula for the hyperbola
The relationship between a point \( P(x, y) \) on the hyperbola, the focus \( S \), and the directrix can be expressed as:
\[ PS = e \cdot PM \]
where \( PS \) is the distance from point \( P \) to the focus \( S \) and \( PM \) is the distance from point \( P \) to the directrix.

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

### Step 4: Calculate \( PM \)
To find the distance \( PM \) from point \( P(x, y) \) to the directrix \( x - y + 3 = 0 \), we use the formula for the distance from a point to a line:
\[ PM = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x - y + 3 = 0 \), we have \( A = 1, B = -1, C = 3 \). Thus,
\[ PM = \frac{|1 \cdot x - 1 \cdot y + 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 3|}{\sqrt{2}} \]

### Step 5: Set up the equation
Using the relationship from Step 2, we have:
\[ \sqrt{(x + 1)^2 + (y - 1)^2} = 3 \cdot \frac{|x - y + 3|}{\sqrt{2}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x + 1)^2 + (y - 1)^2 = 9 \cdot \frac{(x - y + 3)^2}{2} \]

### Step 7: Expand both sides
Expanding the left side:
\[ (x + 1)^2 + (y - 1)^2 = x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 + y^2 + 2x - 2y + 2 \]

Expanding the right side:
\[ 9 \cdot \frac{(x - y + 3)^2}{2} = \frac{9}{2} (x^2 - 2xy + y^2 + 6x - 6y + 9) \]
This simplifies to:
\[ \frac{9}{2}x^2 - 9xy + \frac{9}{2}y^2 + 27x - 27y + \frac{81}{2} \]

### Step 8: Combine and simplify the equation
Setting both sides equal:
\[ x^2 + y^2 + 2x - 2y + 2 = \frac{9}{2}x^2 - 9xy + \frac{9}{2}y^2 + 27x - 27y + \frac{81}{2} \]

### Step 9: Rearranging the equation
Bringing all terms to one side gives:
\[ 0 = \frac{7}{2}x^2 + \frac{7}{2}y^2 - 9xy + 25x - 25y + \frac{77}{2} \]

### Final Equation
Multiplying through by 2 to eliminate the fraction:
\[ 7x^2 + 7y^2 - 18xy + 50x - 50y + 77 = 0 \]

This is the equation of the hyperbola.