Find the equation of the pair of tangents drawn from (-1,2) to the hyperola `2x^(2)-3y^(2)=1`.

Answer

To find the equation of the pair of tangents drawn from the point (-1, 2) to the hyperbola given by the equation \(2x^2 - 3y^2 = 1\), we can follow these steps:

### Step 1: Identify the hyperbola and the point
The hyperbola is given by the equation:
\[ 2x^2 - 3y^2 = 1 \]
The point from which the tangents are drawn is \((-1, 2)\).

### Step 2: Write the equation of the tangent
The general equation of the tangent to the hyperbola \(2x^2 - 3y^2 = 1\) at a point \((x_1, y_1)\) is given by:
\[ T = 2xx_1 - 3yy_1 - 1 = 0 \]
For our point \((-1, 2)\), we substitute \(x_1 = -1\) and \(y_1 = 2\):
\[ T = 2x(-1) - 3y(2) - 1 = -2x - 6y - 1 = 0 \]

### Step 3: Substitute into the hyperbola equation
We need to find the equation of the pair of tangents. The condition for tangents from the point \((-1, 2)\) to the hyperbola is given by:
\[ T^2 = S S_1 \]
where \(S\) is the hyperbola \(2x^2 - 3y^2 - 1 = 0\) and \(S_1\) is obtained by substituting the point \((-1, 2)\) into the hyperbola equation:
\[ S_1 = 2(-1)^2 - 3(2)^2 - 1 = 2 - 12 - 1 = -11 \]

### Step 4: Calculate \(T^2\)
Now, we will calculate \(T^2\):
\[ T = -2x - 6y - 1 \]
So,
\[ T^2 = (-2x - 6y - 1)^2 = 4x^2 + 36y^2 + 1 + 24xy + 4x + 12y \]

### Step 5: Set up the equation
Now we set up the equation:
\[ T^2 = S S_1 \]
Substituting \(S\) and \(S_1\):
\[ 4x^2 + 36y^2 + 1 + 24xy + 4x + 12y = -11(2x^2 - 3y^2 - 1) \]
Expanding the right side:
\[ -22x^2 + 33y^2 + 11 \]

### Step 6: Combine and simplify
Now, we combine both sides:
\[ 4x^2 + 36y^2 + 1 + 24xy + 4x + 12y + 22x^2 - 33y^2 - 11 = 0 \]
This simplifies to:
\[ (4x^2 + 22x^2) + (36y^2 - 33y^2) + 24xy + 4x + 12y - 10 = 0 \]
\[ 26x^2 + 3y^2 + 24xy + 4x + 12y - 10 = 0 \]

### Final Equation
Thus, the equation of the pair of tangents drawn from the point (-1, 2) to the hyperbola \(2x^2 - 3y^2 = 1\) is:
\[ 26x^2 + 3y^2 + 24xy + 4x + 12y - 10 = 0 \]