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. **Identify the equations**:
- The equation of the parabola is \(y^2 = 8x\).
- The equation of the line is \(4x - 3y + 4 = 0\).

2. **Express \(x\) in terms of \(y\) from the line equation**:
\[ 4x = 3y - 4 \implies x = \frac{3y - 4}{4} \]

3. **Substitute \(x\) into the parabola equation**:
Substitute \(x\) in the parabola equation:
\[ y^2 = 8\left(\frac{3y - 4}{4}\right) \]
Simplifying this gives:
\[ y^2 = 6y - 8 \]
Rearranging it leads to:
\[ y^2 - 6y + 8 = 0 \]

4. **Solve the quadratic equation**:
We can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 1\), \(b = -6\), and \(c = 8\):
\[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} \]
Thus, we have:
\[ y = \frac{8}{2} = 4 \quad \text{and} \quad y = \frac{4}{2} = 2 \]

5. **Find corresponding \(x\) values**:
For \(y = 4\):
\[ x = \frac{3(4) - 4}{4} = \frac{12 - 4}{4} = \frac{8}{4} = 2 \]
For \(y = 2\):
\[ x = \frac{3(2) - 4}{4} = \frac{6 - 4}{4} = \frac{2}{4} = \frac{1}{2} \]

6. **Points of intersection**:
The points of intersection are:
- Point A: \((2, 4)\)
- Point B: \(\left(\frac{1}{2}, 2\right)\)

7. **Calculate the midpoint \((\lambda, \mu)\)**:
Using the midpoint formula:
\[ \lambda = \frac{x_1 + x_2}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{\frac{4}{2} + \frac{1}{2}}{2} = \frac{\frac{5}{2}}{2} = \frac{5}{4} \]
\[ \mu = \frac{y_1 + y_2}{2} = \frac{4 + 2}{2} = \frac{6}{2} = 3 \]

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

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