If `y+b=m_1(x+a)& y+b=m_2(x+a)` are two tangents to the parabola `y^2=4ax` then `|m_1m_2|` is equal to:

Answer

To solve the problem, we need to find the value of \(|m_1 m_2|\) given that the lines \(y + b = m_1(x + a)\) and \(y + b = m_2(x + a)\) are tangents to the parabola \(y^2 = 4ax\).

### Step-by-Step Solution:

1. **Identify the Tangent Line Equation**:
The equations of the tangents can be rewritten as:
\[ y = m_1(x + a) - b \quad \text{and} \quad y = m_2(x + a) - b \]
These lines are in the slope-intercept form \(y = mx + c\).

2. **Substituting into the Parabola Equation**:
For a line \(y = mx + c\) to be a tangent to the parabola \(y^2 = 4ax\), it must satisfy the condition:
\[ c = \frac{a}{m} \]
Here, \(c\) is the y-intercept of the line.

3. **Finding the y-intercept**:
From our line equations, the y-intercept \(c\) can be expressed as:
\[ c = -b + ma \]
Thus, we have:
\[ -b + ma = \frac{a}{m} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ ma + b = \frac{a}{m} \]
Multiplying through by \(m\) to eliminate the fraction:
\[ m^2a + bm = a \]
Rearranging this, we get:
\[ m^2a + bm - a = 0 \]

5. **Using the Quadratic Formula**:
This is a quadratic equation in \(m\):
\[ am^2 + bm - a = 0 \]
The product of the roots \(m_1\) and \(m_2\) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by \(\frac{C}{A}\). Here, \(A = a\), \(B = b\), and \(C = -a\).
Therefore, the product of the slopes is:
\[ m_1 m_2 = \frac{-a}{a} = -1 \]

6. **Finding the Absolute Value**:
Finally, we need to find \(|m_1 m_2|\):
\[ |m_1 m_2| = |-1| = 1 \]

### Final Answer:
\[ |m_1 m_2| = 1 \]