Tangents are drawn from (-2,1) to the hyperola `2x^(2)-3y^(2)=6`. Find their equations.

Answer

To find the equations of the tangents drawn from the point (-2, 1) to the hyperbola given by the equation \(2x^2 - 3y^2 = 6\), we will follow these steps:

### Step 1: Convert the hyperbola equation to standard form
The given equation of the hyperbola is:
\[ 2x^2 - 3y^2 = 6 \]
We can divide the entire equation by 6 to convert it into standard form:
\[ \frac{x^2}{3} - \frac{y^2}{2} = 1 \]
Here, we identify \(a^2 = 3\) and \(b^2 = 2\).

### Step 2: Use the tangent equation
The equation of the tangent to the hyperbola in slope form is given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Since the tangents pass through the point (-2, 1), we substitute \(x = -2\) and \(y = 1\) into the tangent equation:
\[ 1 = -2m \pm \sqrt{3m^2 - 2} \]

### Step 3: Solve for \(m\)
We will consider both cases of the equation (using \(+\) and \(-\)) separately.

#### Case 1: Using the positive sign
\[ 1 = -2m + \sqrt{3m^2 - 2} \]
Rearranging gives:
\[ \sqrt{3m^2 - 2} = 1 + 2m \]
Squaring both sides:
\[ 3m^2 - 2 = (1 + 2m)^2 \]
Expanding the right side:
\[ 3m^2 - 2 = 1 + 4m + 4m^2 \]
Rearranging leads to:
\[ 0 = m^2 + 4m + 3 \]

#### Case 2: Using the negative sign
\[ 1 = -2m - \sqrt{3m^2 - 2} \]
Rearranging gives:
\[ \sqrt{3m^2 - 2} = -1 - 2m \]
Since the left side is always non-negative, this case does not yield any valid solutions.

### Step 4: Solve the quadratic equation
From the first case, we have:
\[ m^2 + 4m + 3 = 0 \]
Factoring gives:
\[ (m + 1)(m + 3) = 0 \]
Thus, the solutions for \(m\) are:
\[ m = -1 \quad \text{and} \quad m = -3 \]

### Step 5: Find the equations of the tangents
Now we will substitute the values of \(m\) back into the tangent equation.

#### For \(m = -1\):
\[ y = -x \pm \sqrt{3(-1)^2 - 2} = -x \pm \sqrt{3 - 2} = -x \pm 1 \]
This gives us two tangents:
1. \(y = -x + 1\)
2. \(y = -x - 1\)

#### For \(m = -3\):
\[ y = -3x \pm \sqrt{3(-3)^2 - 2} = -3x \pm \sqrt{27 - 2} = -3x \pm \sqrt{25} \]
This gives us two tangents:
1. \(y = -3x + 5\)
2. \(y = -3x - 5\)

### Final Equations of the Tangents
The equations of the tangents to the hyperbola from the point (-2, 1) are:
1. \(y = -x + 1\) (or \(x + y - 1 = 0\))
2. \(y = -x - 1\) (or \(x + y + 1 = 0\))
3. \(y = -3x + 5\) (or \(3x + y - 5 = 0\))
4. \(y = -3x - 5\) (or \(3x + y + 5 = 0\))