Find the equation of the hyperbola whose focus is `(1,1)` directrix `2x + 2y =1,` and eccentricity `sqrt2.`

Answer

To find the equation of the hyperbola whose focus is at (1,1), directrix is given by the equation \(2x + 2y = 1\), and eccentricity \(e = \sqrt{2}\), we can follow these steps:

### Step 1: Identify the Focus and Directrix
The focus of the hyperbola is given as \(F(1, 1)\) and the directrix is given by the line \(2x + 2y = 1\).

### Step 2: Write the Equation of the Directrix
We can rewrite the equation of the directrix in the standard form:
\[ 2x + 2y - 1 = 0 \]
This can also be expressed as:
\[ A = 2, \quad B = 2, \quad C = -1 \]

### Step 3: Use the Definition of Hyperbola
According to the definition of a hyperbola, for any point \(P(x, y)\) on the hyperbola, the distance from \(P\) to the focus \(F\) is equal to \(e\) times the perpendicular distance from \(P\) to the directrix. Mathematically, this can be expressed as:
\[ SP = e \cdot PM \]
where \(SP\) is the distance from point \(P\) to the focus \(F\) and \(PM\) is the perpendicular distance from point \(P\) to the directrix.

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

### Step 5: Calculate \(PM\)
The perpendicular distance \(PM\) from point \(P(x, y)\) to the directrix \(2x + 2y - 1 = 0\) is given by the formula:
\[ PM = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} = \frac{|2x + 2y - 1|}{\sqrt{2^2 + 2^2}} = \frac{|2x + 2y - 1|}{\sqrt{8}} = \frac{|2x + 2y - 1|}{2\sqrt{2}} \]

### Step 6: Set Up the Equation
Substituting the values of \(SP\) and \(PM\) into the hyperbola definition gives us:
\[ \sqrt{(x - 1)^2 + (y - 1)^2} = \sqrt{2} \cdot \frac{|2x + 2y - 1|}{2\sqrt{2}} \]
This simplifies to:
\[ \sqrt{(x - 1)^2 + (y - 1)^2} = \frac{2x + 2y - 1}{2} \]

### Step 7: Square Both Sides
Squaring both sides to eliminate the square root:
\[ (x - 1)^2 + (y - 1)^2 = \left(\frac{2x + 2y - 1}{2}\right)^2 \]
Expanding both sides:
\[ (x - 1)^2 + (y - 1)^2 = \frac{(2x + 2y - 1)^2}{4} \]

### Step 8: Expand and Rearrange
Expanding the left side:
\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) = x^2 + y^2 - 2x - 2y + 2 \]
Expanding the right side:
\[ \frac{4x^2 + 4y^2 + 1 + 8xy - 4x - 4y}{4} = x^2 + y^2 + \frac{1}{4} + 2xy - x - y \]
Setting both sides equal:
\[ x^2 + y^2 - 2x - 2y + 2 = x^2 + y^2 + \frac{1}{4} + 2xy - x - y \]

### Step 9: Simplify and Collect Like Terms
Rearranging gives:
\[ 0 = 2xy + x + y - 2 - \frac{1}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ 0 = 8xy + 4x + 4y - 8 + 1 \]
Thus, we have:
\[ 8xy + 4x + 4y - 7 = 0 \]

### Final Equation
The equation of the hyperbola is:
\[ 8xy + 4x + 4y - 7 = 0 \]