Find the equation of the hyperbola whose foci are (1,2), `e=sqrt3` and the directrix is 2x+y=1.

Answer

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

### Step 1: Identify the given values
- Foci: \( (1, 2) \)
- Eccentricity: \( e = \sqrt{3} \)
- Directrix: \( 2x + y = 1 \)

### Step 2: Write the general form of the directrix
The directrix can be expressed in the form \( ax + by + c = 0 \). For the given directrix \( 2x + y - 1 = 0 \), we have:
- \( a = 2 \)
- \( b = 1 \)
- \( c = -1 \)

### Step 3: Use the relationship between the foci, eccentricity, and directrix
The relationship for a point \( (h, k) \) on the hyperbola is given by:
\[ \sqrt{(h - x_1)^2 + (k - y_1)^2} = e \cdot \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \]
Substituting the values:
- \( (x_1, y_1) = (1, 2) \)
- \( e = \sqrt{3} \)
- \( a = 2, b = 1, c = -1 \)

The equation becomes:
\[ \sqrt{(h - 1)^2 + (k - 2)^2} = \sqrt{3} \cdot \frac{|2h + k - 1|}{\sqrt{2^2 + 1^2}} \]
Calculating \( \sqrt{2^2 + 1^2} = \sqrt{5} \), we rewrite the equation:
\[ \sqrt{(h - 1)^2 + (k - 2)^2} = \sqrt{3} \cdot \frac{|2h + k - 1|}{\sqrt{5}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - 1)^2 + (k - 2)^2 = \frac{3(2h + k - 1)^2}{5} \]

### Step 5: Expand both sides
Expanding the left side:
\[ (h - 1)^2 + (k - 2)^2 = (h^2 - 2h + 1) + (k^2 - 4k + 4) = h^2 + k^2 - 2h - 4k + 5 \]

Expanding the right side:
\[ \frac{3(2h + k - 1)^2}{5} = \frac{3(4h^2 + 4hk - 4h + k^2 - 2k + 1)}{5} = \frac{12h^2 + 12hk - 12h + 3k^2 - 6k + 3}{5} \]

### Step 6: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 5(h^2 + k^2 - 2h - 4k + 5) = 12h^2 + 12hk - 12h + 3k^2 - 6k + 3 \]

### Step 7: Rearrange the equation
Rearranging all terms to one side results in:
\[ 5h^2 + 5k^2 - 10h - 20k + 25 - 12h^2 - 12hk + 12h - 3k^2 + 6k - 3 = 0 \]
Combining like terms:
\[ -7h^2 + 2k^2 - 12hk + 2h - 14k + 22 = 0 \]

### Step 8: Substitute \( h \) and \( k \) with \( x \) and \( y \)
Let \( h = x \) and \( k = y \):
\[ -7x^2 + 2y^2 - 12xy + 2x - 14y + 22 = 0 \]
Multiplying through by -1 gives the final equation:
\[ 7x^2 - 2y^2 + 12xy - 2x + 14y - 22 = 0 \]

### Final Equation
The equation of the hyperbola is:
\[ 7x^2 - 2y^2 + 12xy - 2x + 14y - 22 = 0 \]