A chord is drawn through the focus of the parabola $$y^2 = 6x$$ such that its distance from the vertex of this parabola is $$\frac{\sqrt{5}}{2}$$, then its slope can be:

The given parabola is $$ y^2 = 6x $$. Comparing this with the standard form $$ y^2 = 4ax $$, we find $$ 4a = 6 $$, so $$ a = \frac{3}{2} $$. The focus of the parabola is at $$ \left( \frac{3}{2}, 0 \right) $$, and the vertex is at $$ (0, 0) $$. A chord is drawn through the focus $$ \left( \frac{3}{2}, 0 \right) $$ with slope $$ m $$. The equation of this chord is: $$ y - 0 = m \left( x - \frac{3}{2} \right) $$ which simplifies to: $$ y = m \left( x - \frac{3}{2} \right) $$ Rewriting this in the standard form $$ ax + by + c = 0 $$: $$ m x - y - \frac{3m}{2} = 0 $$ Here, $$ a = m $$, $$ b = -1 $$, and $$ c = -\frac{3m}{2} $$. The perpendicular distance from the vertex $$ (0, 0) $$ to this line is given by the formula: $$ \text{distance} = \frac{|a \cdot 0 + b \cdot 0 + c|}{\sqrt{a^2 + b^2}} = \frac{| -\frac{3m}{2} |}{\sqrt{m^2 + (-1)^2}} = \frac{\frac{3|m|}{2}}{\sqrt{m^2 + 1}} = \frac{3|m|}{2\sqrt{m^2 + 1}} $$ This distance is given as $$ \frac{\sqrt{5}}{2} $$. So: $$ \frac{3|m|}{2\sqrt{m^2 + 1}} = \frac{\sqrt{5}}{2} $$ Multiplying both sides by 2: $$ \frac{3|m|}{\sqrt{m^2 + 1}} = \sqrt{5} $$ Since the options are positive, assume $$ m > 0 $$ (the solution will account for both signs later). Thus, $$ |m| = m $$: $$ \frac{3m}{\sqrt{m^2 + 1}} = \sqrt{5} $$ Squaring both sides to eliminate the square root: $$ \left( \frac{3m}{\sqrt{m^2 + 1}} \right)^2 = (\sqrt{5})^2 $$ $$ \frac{9m^2}{m^2 + 1} = 5 $$ Cross-multiplying: $$ 9m^2 = 5(m^2 + 1) $$ $$ 9m^2 = 5m^2 + 5 $$ Bringing all terms to one side: $$ 9m^2 - 5m^2 - 5 = 0 $$ $$ 4m^2 - 5 = 0 $$ $$ 4m^2 = 5 $$ $$ m^2 = \frac{5}{4} $$ $$ m = \pm \frac{\sqrt{5}}{2} $$ The slopes are $$ \frac{\sqrt{5}}{2} $$ and $$ -\frac{\sqrt{5}}{2} $$. Now, verify these satisfy the original distance condition. For $$ m = \frac{\sqrt{5}}{2} $$: $$ \text{distance} = \frac{3 \left| \frac{\sqrt{5}}{2} \right|}{2 \sqrt{ \left( \frac{\sqrt{5}}{2} \right)^2 + 1 }} = \frac{3 \cdot \frac{\sqrt{5}}{2}}{2 \sqrt{ \frac{5}{4} + 1 }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \sqrt{ \frac{5}{4} + \frac{4}{4} }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \sqrt{ \frac{9}{4} }} = \frac{ \frac{3\sqrt{5}}{2} }{2 \cdot \frac{3}{2} } = \frac{ \frac{3\sqrt{5}}{2} }{3} = \frac{3\sqrt{5}}{2} \cdot \frac{1}{3} = \frac{\sqrt{5}}{2} $$ For $$ m = -\frac{\sqrt{5}}{2} $$: $$ \text{distance} = \frac{3 \left| -\frac{\sqrt{5}}{2} \right|}{2 \sqrt{ \left( -\frac{\sqrt{5}}{2} \right)^2 + 1 }} = \frac{3 \cdot \frac{\sqrt{5}}{2}}{2 \sqrt{ \frac{5}{4} + 1 }} = \text{same as above} = \frac{\sqrt{5}}{2} $$ Both slopes satisfy the distance condition. Now, check if the chord intersects the parabola. For $$ m = \frac{\sqrt{5}}{2} $$, substitute into the parabola $$ y^2 = 6x $$: $$ y = \frac{\sqrt{5}}{2} \left( x - \frac{3}{2} \right) $$ $$ \left[ \frac{\sqrt{5}}{2} \left( x - \frac{3}{2} \right) \right]^2 = 6x $$ $$ \frac{5}{4} \left( x - \frac{3}{2} \right)^2 = 6x $$ Multiply by 4: $$ 5 \left( x - \frac{3}{2} \right)^2 = 24x $$ $$ 5 \left( x^2 - 3x + \frac{9}{4} \right) = 24x $$ $$ 5x^2 - 15x + \frac{45}{4} = 24x $$ $$ 5x^2 - 15x - 24x + \frac{45}{4} = 0 $$ $$ 5x^2 - 39x + \frac{45}{4} = 0 $$ Multiply by 4: $$ 20x^2 - 156x + 45 = 0 $$ Discriminant: $$ d = (-156)^2 - 4 \cdot 20 \cdot 45 = 24336 - 3600 = 20736 $$ $$ \sqrt{20736} = 144 \quad (144^2 = 20736) $$ Real roots exist. Similarly, for $$ m = -\frac{\sqrt{5}}{2} $$, the chord intersects the parabola. Among the options, $$ \frac{\sqrt{5}}{2} $$ is option A. The other options do not satisfy the distance condition: - Option B: $$ m = \frac{2}{\sqrt{3}} $$ gives distance $$ \approx 1.134 \neq \frac{\sqrt{5}}{2} \approx 1.118 $$ - Option C: $$ m = \frac{\sqrt{3}}{2} $$ gives distance $$ \approx 0.982 \neq \frac{\sqrt{5}}{2} $$ - Option D: $$ m = \frac{2}{\sqrt{5}} $$ gives distance $$ = 1 \neq \frac{\sqrt{5}}{2} $$ Hence, the slope can be $$ \frac{\sqrt{5}}{2} $$, which is option A. So, the answer is Option A.