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 \(2x^2 - 3y^2 = 6\), we can follow these steps:

### Step 1: Rewrite the equation of the hyperbola in standard form
The equation of the hyperbola is given as:
\[ 2x^2 - 3y^2 = 6 \]
Dividing the entire equation by 6, we get:
\[ \frac{x^2}{3} - \frac{y^2}{2} = 1 \]
This shows that \(a^2 = 3\) and \(b^2 = 2\).

### Step 2: Identify the slopes of the tangents
The general formula for the slope \(m\) of the tangents to the hyperbola is given by:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
Substituting \(a^2\) and \(b^2\):
\[ y = mx \pm \sqrt{3m^2 - 2} \]

### Step 3: Substitute the point (-2, 1) into the tangent equation
We substitute \(x = -2\) and \(y = 1\) into the tangent equation:
\[ 1 = m(-2) \pm \sqrt{3m^2 - 2} \]
This simplifies to:
\[ 1 = -2m \pm \sqrt{3m^2 - 2} \]

### Step 4: Solve for \(m\)
We will consider both cases for the plus and minus sign.

#### Case 1: \(1 = -2m + \sqrt{3m^2 - 2}\)
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 gives:
\[ 3m^2 - 4m^2 - 4m - 3 = 0 \implies -m^2 - 4m - 3 = 0 \implies m^2 + 4m + 3 = 0 \]
Factoring:
\[ (m + 3)(m + 1) = 0 \]
Thus, \(m = -3\) or \(m = -1\).

#### Case 2: \(1 = -2m - \sqrt{3m^2 - 2}\)
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 5: Find the equations of the tangents
Using the slopes \(m = -3\) and \(m = -1\):

1. For \(m = -1\):
\[ y = -x + \sqrt{3(-1)^2 - 2} = -x + 1 \]
The equation is:
\[ y = -x + 1 \]

2. For \(m = -3\):
\[ y = -3x + \sqrt{3(-3)^2 - 2} = -3x + 5 \]
The equation is:
\[ y = -3x + 5 \]

### Final Result
The equations of the tangents are:
\[ y = -x + 1 \quad \text{and} \quad y = -3x + 5 \]