Find the equation of hyperbola whose equation of directrix is x+2y=1, focus is (-1,-1) and eccentricity is `sqrt(2)`.

Answer

To find the equation of the hyperbola with the given conditions, we will use the focal-directrix property of conics. Let's break down the solution step by step.

### Step 1: Identify the Given Information
- Focus (F) = (-1, -1)
- Directrix (D): x + 2y = 1
- Eccentricity (e) = √2

### Step 2: Rewrite the Directrix Equation
We can rewrite the directrix in the standard form:
\[ x + 2y - 1 = 0 \]
Here, we have:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -1 \)

### Step 3: Use the Focal-Directrix Property
According to the focal-directrix property, for any point \( P(x, y) \) on the hyperbola, the following relationship holds:
\[ PF = e \cdot PM \]
Where:
- \( PF \) is the distance from the point \( P \) to the focus \( F \).
- \( PM \) is the perpendicular distance from the point \( P \) to the directrix \( D \).

### Step 4: Calculate the Distance \( PF \)
Using the distance formula, the distance \( PF \) from point \( P(x, y) \) to the focus \( F(-1, -1) \) is given by:
\[ PF = \sqrt{(x + 1)^2 + (y + 1)^2} \]

### Step 5: Calculate the Perpendicular Distance \( PM \)
The formula for the perpendicular distance from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is:
\[ PM = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Substituting the values for our directrix:
\[ PM = \frac{|1 \cdot x + 2 \cdot y - 1|}{\sqrt{1^2 + 2^2}} = \frac{|x + 2y - 1|}{\sqrt{5}} \]

### Step 6: Set Up the Equation
From the focal-directrix property, we have:
\[ PF = e \cdot PM \]
Substituting the expressions we found:
\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{2} \cdot \frac{|x + 2y - 1|}{\sqrt{5}} \]

### Step 7: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ (x + 1)^2 + (y + 1)^2 = \frac{2}{5}(x + 2y - 1)^2 \]

### Step 8: 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:
\[ \frac{2}{5}(x + 2y - 1)^2 = \frac{2}{5}(x^2 + 4y^2 + 4xy - 2x - 4y + 1) \]

### Step 9: Combine and Rearrange
Combining both sides and simplifying leads to an equation in standard form. After simplification, we will get:
\[ 3x^2 - 3y^2 + 14x + 18y + 8 = 0 \]

### Final Equation
Thus, the equation of the hyperbola is:
\[ 3x^2 - 3y^2 + 14x + 18y + 8 = 0 \]