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: Rewrite the Hyperbola Equation
First, we rewrite the hyperbola equation in standard form. We divide the entire equation by 6:
\[
\frac{x^2}{3} - \frac{y^2}{2} = 1
\]
Here, we can identify \(a^2 = 3\) and \(b^2 = 2\).
### Step 2: Use the Tangent Equation Formula
The equation of the tangent to the hyperbola from an external point \((x_0, y_0)\) can be given by the formula:
\[
\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1
\]
Substituting \(x_0 = -2\), \(y_0 = -1\), \(a^2 = 3\), and \(b^2 = 2\):
\[
\frac{x(-2)}{3} - \frac{y(-1)}{2} = 1
\]
### Step 3: Simplify the Tangent Equation
Now, we simplify the equation:
\[
-\frac{2x}{3} + \frac{y}{2} = 1
\]
To eliminate the fractions, we can multiply through by 6 (the least common multiple of 3 and 2):
\[
-4x + 3y = 6
\]
Rearranging gives us:
\[
4x - 3y + 6 = 0 \quad \text{(Equation 1)}
\]
### Step 4: Find the Slope of the Tangents
To find the slopes of the tangents, we can use the general form of the tangent line to the hyperbola:
\[
y = mx + c
\]
Substituting this into the hyperbola equation, we can derive a quadratic equation in terms of \(m\). However, we can also directly use the point-slope form.
### Step 5: Substitute Point into the Tangent Equation
Since the point \((-2, -1)\) lies on the tangents, we can substitute \(y = mx + c\) into the hyperbola equation and find the conditions for tangency (discriminant = 0).
After substituting and simplifying, we will arrive at a quadratic equation in \(m\):
\[
m^2 - 4m + 3 = 0
\]
### Step 6: Solve for \(m\)
Now, we solve the quadratic equation:
\[
m^2 - 4m + 3 = 0
\]
Factoring gives:
\[
(m - 3)(m - 1) = 0
\]
Thus, the slopes \(m\) are:
\[
m = 3 \quad \text{and} \quad m = 1
\]
### Step 7: Write the Tangent Equations
Using the point-slope form \(y - y_1 = m(x - x_1)\):
1. For \(m = 3\):
\[
y + 1 = 3(x + 2) \implies y = 3x + 6 - 1 \implies 3x - y + 5 = 0
\]
2. For \(m = 1\):
\[
y + 1 = 1(x + 2) \implies y = x + 2 - 1 \implies x - y + 1 = 0
\]
### Final Answer
Thus, the equations of the tangents from the point \((-2, -1)\) to the hyperbola \(2x^2 - 3y^2 = 6\) are:
1. \(3x - y + 5 = 0\)
2. \(x - y + 1 = 0\)
