A line parallel to the straight line $$2x - y = 0$$ is tangent to the hyperbola $$\frac{x^2}{4} - \frac{y^2}{2} = 1$$ at the point $$(x_1, y_1)$$. Then $$x_1^2 + 5y_1^2$$ is equal to:

We are given the hyperbola $$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$$ whose standard form is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ with $$a^{2}=4$$ and $$b^{2}=2.$$

For this hyperbola the point-form of the tangent at the point $$(x_{1},y_{1})$$ on the curve is, by the standard formula,

$$\frac{x\,x_{1}}{a^{2}}-\frac{y\,y_{1}}{b^{2}}=1.$$

Substituting $$a^{2}=4$$ and $$b^{2}=2$$ we obtain the required tangent equation:

$$\frac{x\,x_{1}}{4}-\frac{y\,y_{1}}{2}=1.$$

Now we rearrange this equation in the slope-intercept form so that the slope can be compared with the given line. First multiply every term by $$4$$ to clear the denominators:

$$x\,x_{1}-2y\,y_{1}=4.$$

Next isolate the $$y$$-term:

$$-2y\,y_{1}=4-x\,x_{1}.$$

Divide both sides by $$-2y_{1}:$$

$$y=\frac{x\,x_{1}}{2y_{1}}-\frac{4}{2y_{1}}.$$

Simplifying, we get

$$y=\frac{x_{1}}{2y_{1}}\,x-\frac{2}{y_{1}}.$$

Thus the slope $$m$$ of the tangent line is

$$m=\frac{x_{1}}{2y_{1}}.$$

We are told that this tangent is parallel to the straight line $$2x-y=0.$$ Writing that line in slope-intercept form gives $$y=2x,$$ whose slope is $$m=2.$$ Because parallel lines have equal slopes, we equate the two slopes:

$$\frac{x_{1}}{2y_{1}}=2.$$

Multiplying both sides by $$2y_{1}$$ yields

$$x_{1}=4y_{1}.$$

Since $$(x_{1},y_{1})$$ lies on the given hyperbola, it must satisfy its equation. Substituting $$x_{1}=4y_{1}$$ into

$$\frac{x_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1$$

gives

$$\frac{(4y_{1})^{2}}{4}-\frac{y_{1}^{2}}{2}=1.$$

Compute the first term:

$$\frac{16y_{1}^{2}}{4}-\frac{y_{1}^{2}}{2}=1,$$

which simplifies to

$$4y_{1}^{2}-\frac{y_{1}^{2}}{2}=1.$$

Write $$4y_{1}^{2}$$ with a denominator of $$2$$ so we can combine like terms:

$$\frac{8y_{1}^{2}}{2}-\frac{y_{1}^{2}}{2}=1.$$

This gives

$$\frac{7y_{1}^{2}}{2}=1.$$

Multiplying both sides by $$\frac{2}{7}$$ produces

$$y_{1}^{2}=\frac{2}{7}.$$

Using $$x_{1}=4y_{1},$$ we have

$$x_{1}^{2}=16y_{1}^{2}=16\left(\frac{2}{7}\right)=\frac{32}{7}.$$

Now we compute the required expression $$x_{1}^{2}+5y_{1}^{2}:$$

$$x_{1}^{2}+5y_{1}^{2}=\frac{32}{7}+5\left(\frac{2}{7}\right)=\frac{32}{7}+\frac{10}{7}=\frac{42}{7}=6.$$

Hence, the correct answer is Option A.