If `2x +3y = alpha, x -y = beta` and `kx +15y = r` are 3 concurrent normal of parabola `y^(2) = lambda x` then value of k is (a) 3 (b) 4 (c) 5 (d) 7

Options

A. 3
B. 4
C. 5
D. 7

Explanation

To solve the problem, we need to find the value of \( k \) such that the three given lines are concurrent normals to the parabola \( y^2 = \lambda x \).

### Step-by-step Solution:

1. **Identify the equations of the normals**:
- The first normal is given by \( 2x + 3y = \alpha \).
- The second normal is given by \( x - y = \beta \).
- The third normal is given by \( kx + 15y = r \).

2. **Find the slope of the first normal**:
- Rearranging the first equation:
\[ 3y = -2x + \alpha \implies y = -\frac{2}{3}x + \frac{\alpha}{3} \]
- The slope \( m_1 \) of the first normal is:
\[ m_1 = -\frac{2}{3} \]

3. **Find the slope of the second normal**:
- Rearranging the second equation:
\[ y = x - \beta \]
- The slope \( m_2 \) of the second normal is:
\[ m_2 = 1 \]

4. **Find the slope of the third normal**:
- Rearranging the third equation:
\[ 15y = -kx + r \implies y = -\frac{k}{15}x + \frac{r}{15} \]
- The slope \( m_3 \) of the third normal is:
\[ m_3 = -\frac{k}{15} \]

5. **Set up the equation for concurrency**:
- For the three normals to be concurrent, the sum of their slopes must equal zero:
\[ m_1 + m_2 + m_3 = 0 \]
- Substituting the values of the slopes:
\[ -\frac{2}{3} + 1 - \frac{k}{15} = 0 \]

6. **Simplify the equation**:
- Combine the terms:
\[ 1 - \frac{2}{3} = \frac{1}{3} \]
- Thus, we have:
\[ \frac{1}{3} - \frac{k}{15} = 0 \]

7. **Solve for \( k \)**:
- Rearranging gives:
\[ \frac{k}{15} = \frac{1}{3} \]
- Cross-multiplying:
\[ k = 15 \times \frac{1}{3} = 5 \]

### Conclusion:
The value of \( k \) is \( 5 \).