The mid-point of the chord intercepted on the line `4x-3y+4=0` by the parabola `y^2=8x" is "(lambda,mu)`, thus the value of `(4lambda+mu)` is

Options

A. 2
B. 4
C. 8
D. 6

Answer

To solve the problem, we need to find the midpoint of the chord intercepted on the line \(4x - 3y + 4 = 0\) by the parabola \(y^2 = 8x\), and then calculate the value of \(4\lambda + \mu\).

### Step-by-Step Solution:

1. **Find the equation of the line in terms of \(y\)**:
The line equation is given as:
\[ 4x - 3y + 4 = 0 \]
Rearranging gives:
\[ 3y = 4x + 4 \quad \Rightarrow \quad y = \frac{4x + 4}{3} \]

2. **Substitute \(y\) into the parabola equation**:
The parabola is given by:
\[ y^2 = 8x \]
Substitute \(y = \frac{4x + 4}{3}\) into the parabola equation:
\[ \left(\frac{4x + 4}{3}\right)^2 = 8x \]
Expanding the left side:
\[ \frac{(4x + 4)^2}{9} = 8x \quad \Rightarrow \quad \frac{16x^2 + 32x + 16}{9} = 8x \]
Multiplying through by 9 to eliminate the fraction:
\[ 16x^2 + 32x + 16 = 72x \]
Rearranging gives:
\[ 16x^2 - 40x + 16 = 0 \]

3. **Simplify the quadratic equation**:
Dividing the entire equation by 8:
\[ 2x^2 - 5x + 2 = 0 \]

4. **Find the roots of the quadratic equation**:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 2\), \(b = -5\), and \(c = 2\):
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us:
\[ x_1 = \frac{8}{4} = 2 \quad \text{and} \quad x_2 = \frac{2}{4} = \frac{1}{2} \]

5. **Find the corresponding \(y\) values**:
For \(x_1 = 2\):
\[ y = \frac{4(2) + 4}{3} = \frac{8 + 4}{3} = 4 \]
For \(x_2 = \frac{1}{2}\):
\[ y = \frac{4\left(\frac{1}{2}\right) + 4}{3} = \frac{2 + 4}{3} = 2 \]

6. **Determine the coordinates of points A and B**:
The points are:
\[ A\left(\frac{1}{2}, 2\right) \quad \text{and} \quad B(2, 4) \]

7. **Find the midpoint \((\lambda, \mu)\)**:
The midpoint formula is:
\[ \left(\lambda, \mu\right) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Thus,
\[ \lambda = \frac{\frac{1}{2} + 2}{2} = \frac{\frac{1}{2} + \frac{4}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4} \]
\[ \mu = \frac{2 + 4}{2} = \frac{6}{2} = 3 \]

8. **Calculate \(4\lambda + \mu\)**:
\[ 4\lambda + \mu = 4 \cdot \frac{5}{4} + 3 = 5 + 3 = 8 \]

### Final Answer:
The value of \(4\lambda + \mu\) is \(8\).