If l and m are variable real number such that `5l^(2)+6m^(2)-4lm+3l=0`, then the variable line lx+my=1 always touches a fixed parabola, whose axes is parallel to the x-axis.
The directrix of the parabola is

Options

A. 6x+7=0
B. 4x+11=0
C. 3x+11=0
D. none of these

Explanation

To solve the problem step by step, we will analyze the given equation and derive the equation of the directrix of the parabola.

### Step 1: Understand the given equation
We start with the equation:
\[ 5l^2 + 6m^2 - 4lm + 3l = 0 \]
This represents a quadratic equation in terms of \( l \) and \( m \).

### Step 2: Identify the form of the parabola
The problem states that the variable line \( lx + my = 1 \) always touches a fixed parabola whose axis is parallel to the x-axis. The general equation of such a parabola can be written as:
\[ (y - a)^2 = 4b(x - c) \]
where \( (c, a) \) is the vertex of the parabola.

### Step 3: Rewrite the line equation
The line can be expressed in slope-intercept form:
\[ y = -\frac{l}{m}x + \frac{1}{m} \]
The slope of the line is \( -\frac{l}{m} \).

### Step 4: Condition for tangency
For the line to be tangent to the parabola, we need to set up the condition:
\[ (y - a) = -\frac{l}{m}(x - c) + \frac{1 - am - lc}{m} \]
This means we need to ensure that the quadratic formed by substituting \( y \) into the parabola's equation has a discriminant equal to zero.

### Step 5: Set up the discriminant condition
The condition for tangency leads to:
\[ Cl^2 - Bm^2 + Alm = 1 \]
where \( A, B, C \) are coefficients derived from the comparison of the two equations.

### Step 6: Compare coefficients
From the given equation \( 5l^2 + 6m^2 - 4lm + 3l = 0 \), we can compare coefficients:
- \( C = 5 \)
- \( B = 6 \)
- \( A = -4 \)

### Step 7: Solve for \( a, b, c \)
Using the relationships:
\[ \frac{C}{5} = \frac{-B}{6} = \frac{A}{-4} = \frac{-1}{3} \]
We can derive:
- \( C = -\frac{5}{3} \)
- \( B = 2 \)
- \( A = \frac{4}{3} \)

### Step 8: Write the equation of the parabola
Substituting \( a, b, c \) into the parabola equation:
\[ (y - \frac{4}{3})^2 = 4(2)(x + \frac{5}{3}) \]

### Step 9: Find the focus and directrix
The focus of the parabola is at:
\[ \left( \frac{1}{3}, \frac{4}{3} \right) \]
The directrix of the parabola is given by the equation:
\[ x + a = 0 \]
where \( a = 2 \).

### Step 10: Finalize the directrix equation
Thus, the equation of the directrix is:
\[ x + \frac{5}{3} = 0 \]
or, rearranging gives:
\[ x = -\frac{5}{3} \]

### Conclusion
The directrix of the parabola is:
\[ x = -\frac{5}{3} \]