A chord is drawn through the focus of the parabola `y^(2)=6x` such that its distance from the vertex of this parabola is `(sqrt5)/(2)`, then its slope can be :

Options

A. `(sqrt5)/(2)`
B. `(sqrt3)/(2)`
C. `(2)/(sqrt5)`
D. `(2)/(sqrt3)`

Answer

To solve the problem, we need to find the slope of a chord that passes through the focus of the parabola \( y^2 = 6x \) and is at a distance of \( \frac{\sqrt{5}}{2} \) from the vertex.

### Step-by-step Solution:

1. **Identify the parameters of the parabola:**
The given parabola is \( y^2 = 6x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( 4a = 6 \). Therefore, we find:
\[ a = \frac{6}{4} = \frac{3}{2} \]

2. **Determine the focus of the parabola:**
The focus of the parabola \( y^2 = 4ax \) is located at the point \( (a, 0) \). Substituting the value of \( a \):
\[ \text{Focus} = \left( \frac{3}{2}, 0 \right) \]

3. **Set up the equation of the chord:**
A chord passing through the focus can be expressed in the slope-intercept form as:
\[ y = mx + c \]
Since it passes through the focus \( \left( \frac{3}{2}, 0 \right) \), we can substitute these coordinates into the equation to find \( c \):
\[ 0 = m \cdot \frac{3}{2} + c \implies c = -\frac{3m}{2} \]
Thus, the equation of the chord becomes:
\[ y = mx - \frac{3m}{2} \]

4. **Calculate the distance from the vertex to the chord:**
The distance \( D \) from the vertex \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( y = mx - \frac{3m}{2} \), we can rewrite it as:
\[ mx - y - \frac{3m}{2} = 0 \]
Here, \( A = m \), \( B = -1 \), and \( C = -\frac{3m}{2} \). The distance from the vertex \( (0, 0) \) is:
\[ D = \frac{|0 + 0 - \frac{3m}{2}|}{\sqrt{m^2 + 1}} = \frac{\frac{3|m|}{2}}{\sqrt{m^2 + 1}} \]

5. **Set the distance equal to \( \frac{\sqrt{5}}{2} \):**
We set the distance equal to \( \frac{\sqrt{5}}{2} \):
\[ \frac{\frac{3|m|}{2}}{\sqrt{m^2 + 1}} = \frac{\sqrt{5}}{2} \]
Multiplying both sides by \( 2 \) gives:
\[ \frac{3|m|}{\sqrt{m^2 + 1}} = \sqrt{5} \]

6. **Square both sides to eliminate the square root:**
\[ 9m^2 = 5(m^2 + 1) \]
Expanding the right side:
\[ 9m^2 = 5m^2 + 5 \]
Rearranging gives:
\[ 4m^2 = 5 \implies m^2 = \frac{5}{4} \]

7. **Find the slope \( m \):**
Taking the square root of both sides, we find:
\[ m = \pm \frac{\sqrt{5}}{2} \]

### Final Answer:
The possible slopes of the chord are:
\[ m = \frac{\sqrt{5}}{2} \quad \text{or} \quad m = -\frac{\sqrt{5}}{2} \]