If `y+3= m_1 (x+2)` and `y+3= m_2 (x+2)` are two tangents to the parabola `y^2 = 8x`, then

Options

A. `m_1 + m_2 =0`
B. `m_1m_2=-1`
C. `m_1m_2=1`
D. none

Answer

To solve the problem, we need to find the values of \( m_1 \) and \( m_2 \) for the tangents to the parabola \( y^2 = 8x \) given by the equations \( y + 3 = m_1(x + 2) \) and \( y + 3 = m_2(x + 2) \).

### Step-by-Step Solution:

1. **Identify the Point of Tangency**:
The tangents are given in the form \( y + 3 = m(x + 2) \). This indicates that both tangents pass through the point \( (-2, -3) \).

2. **Equation of the Parabola**:
The equation of the parabola is \( y^2 = 8x \). We can rewrite it in the standard form to identify the parameters:
\[ y^2 = 4ax \quad \text{where } a = 2. \]

3. **Equation of the Tangent to the Parabola**:
The equation of the tangent to the parabola \( y^2 = 8x \) at a point \( (x_0, y_0) \) is given by:
\[ yy_0 = 4a(x + x_0). \]
In our case, since the tangents pass through the point \( (-2, -3) \), we can substitute \( x_0 = -2 \) and \( y_0 = -3 \).

4. **Substituting into the Tangent Equation**:
Substitute \( a = 2 \) into the tangent equation:
\[ y(-3) = 4(2)(x + (-2)). \]
Simplifying this gives:
\[ -3y = 8(x - 2). \]
Rearranging leads to:
\[ 3y = -8x + 16 \quad \Rightarrow \quad y = -\frac{8}{3}x + \frac{16}{3}. \]

5. **Finding the Slopes**:
The slope of the tangent line is \( m = -\frac{8}{3} \). Since we have two tangents, we denote the slopes as \( m_1 \) and \( m_2 \).

6. **Using the Condition for Tangents**:
For the tangents to the parabola, we can use the condition that the sum and product of the slopes of the tangents can be derived from the quadratic formed by the slopes:
\[ m_1 + m_2 = -\frac{b}{a} \quad \text{and} \quad m_1 m_2 = \frac{c}{a}. \]
Here, we can derive the quadratic equation from the tangents.

7. **Forming the Quadratic Equation**:
From the previous steps, we can form the quadratic equation:
\[ 2m^2 - 3m - 2 = 0. \]

8. **Finding Roots of the Quadratic**:
Using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}. \]
This simplifies to:
\[ m = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}. \]
Thus, the roots are:
\[ m_1 = 2 \quad \text{and} \quad m_2 = -\frac{1}{2}. \]

9. **Conclusion**:
The values of \( m_1 \) and \( m_2 \) are:
\[ m_1 + m_2 = \frac{3}{2} \quad \text{and} \quad m_1 m_2 = -1. \]