If `m_1,m_2` are the slopes of the two tangents that are drawn from (2,3) to the parabola `y^2=4x` , then the value of `1/m_1+1/m_2` is

Options

A. -3
B. 3
C. `2/3`
D. `3/2`

Answer

To solve the problem, we need to find the slopes \( m_1 \) and \( m_2 \) of the tangents drawn from the point \( (2, 3) \) to the parabola \( y^2 = 4x \), and then calculate \( \frac{1}{m_1} + \frac{1}{m_2} \).

### Step-by-Step Solution:

1. **Equation of the Tangent to the Parabola**:
The equation of the tangent to the parabola \( y^2 = 4x \) at a point with slope \( m \) is given by:
\[ y = mx + \frac{1}{m} \]

2. **Substituting the Point (2, 3)**:
Since the tangent passes through the point \( (2, 3) \), we substitute \( x = 2 \) and \( y = 3 \) into the tangent equation:
\[ 3 = 2m + \frac{1}{m} \]

3. **Multiplying Through by \( m \)**:
To eliminate the fraction, multiply the entire equation by \( m \):
\[ 3m = 2m^2 + 1 \]

4. **Rearranging to Form a Quadratic Equation**:
Rearranging gives us:
\[ 2m^2 - 3m + 1 = 0 \]

5. **Finding the Roots of the Quadratic**:
We can find the roots \( m_1 \) and \( m_2 \) using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -3 \), and \( c = 1 \):
\[ m = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us the roots:
\[ m_1 = \frac{4}{4} = 1 \quad \text{and} \quad m_2 = \frac{2}{4} = \frac{1}{2} \]

6. **Calculating \( \frac{1}{m_1} + \frac{1}{m_2} \)**:
Now we can calculate:
\[ \frac{1}{m_1} + \frac{1}{m_2} = \frac{1}{1} + \frac{1}{\frac{1}{2}} = 1 + 2 = 3 \]

### Final Answer:
The value of \( \frac{1}{m_1} + \frac{1}{m_2} \) is \( 3 \).