From the following information, find the equation of the hyperbola. Focus `(-2,1) ,` Directrix: `2x - 3y + 1 =0, e = (2)/(sqrt3)`

Answer

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

### Step 1: Identify the Given Information
- Focus \( F(-2, 1) \)
- Directrix: \( 2x - 3y + 1 = 0 \)
- Eccentricity \( e = \frac{2}{\sqrt{3}} \)

### Step 2: Write the Distance Formula
Let \( P(x, y) \) be any point on the hyperbola. The distance from point \( P \) to the focus \( F \) is given by:
\[ FP = \sqrt{(x + 2)^2 + (y - 1)^2} \]
The distance from point \( P \) to the directrix can be calculated using the formula for the distance from a point to a line. For the line \( 2x - 3y + 1 = 0 \), the distance \( d \) from point \( P(x, y) \) is:
\[ d = \frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = \frac{|2x - 3y + 1|}{\sqrt{13}} \]

### Step 3: Set Up the Relationship
According to the definition of a hyperbola, the distance from the focus to a point \( P \) is equal to the eccentricity times the distance from the point \( P \) to the directrix:
\[ FP = e \cdot d \]
Substituting the expressions we found:
\[ \sqrt{(x + 2)^2 + (y - 1)^2} = \frac{2}{\sqrt{3}} \cdot \frac{|2x - 3y + 1|}{\sqrt{13}} \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides:
\[ (x + 2)^2 + (y - 1)^2 = \left(\frac{2}{\sqrt{3}} \cdot \frac{|2x - 3y + 1|}{\sqrt{13}}\right)^2 \]
This simplifies to:
\[ (x + 2)^2 + (y - 1)^2 = \frac{4}{3} \cdot \frac{(2x - 3y + 1)^2}{13} \]

### Step 5: Clear the Denominator
Multiply both sides by \( 39 \) (which is \( 3 \times 13 \)):
\[ 39((x + 2)^2 + (y - 1)^2) = 4(2x - 3y + 1)^2 \]

### Step 6: Expand Both Sides
Expanding the left-hand side:
\[ 39((x^2 + 4x + 4) + (y^2 - 2y + 1)) = 39x^2 + 156x + 156 + 39y^2 - 78y \]
Expanding the right-hand side:
\[ 4(4x^2 - 12xy + 9y^2 + 8x - 6y + 1) = 16x^2 - 48xy + 36y^2 + 32x - 24y + 4 \]

### Step 7: Rearrange the Equation
Combine all terms to one side:
\[ 39x^2 + 39y^2 + 156x - 78y + 156 - (16x^2 - 48xy + 36y^2 + 32x - 24y + 4) = 0 \]
This simplifies to:
\[ (39 - 16)x^2 + (39 - 36)y^2 + (156 - 32)x + (-78 + 24)y + (156 - 4) = 0 \]
Which gives:
\[ 23x^2 + 3y^2 + 124x - 54y + 152 = 0 \]

### Final Equation of the Hyperbola
Thus, the equation of the hyperbola is:
\[ 23x^2 + 3y^2 + 124x - 54y + 152 = 0 \]