If the eccentricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is:

We want the equation of the tangent to a standard hyperbola that passes through the point $$(4,6)$$ and whose eccentricity is given as $$e = 2.$$

For a standard hyperbola centred at the origin with its transverse axis along the $$x$$-axis, the equation is taken as

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

The eccentricity formula for this hyperbola is first stated:

$$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$$

We are told that $$e = 2,$$ so we substitute to obtain

$$2 = \sqrt{1 + \frac{b^{2}}{a^{2}}}.$$

Squaring both sides, we get

$$4 = 1 + \frac{b^{2}}{a^{2}}.$$

Now subtract $$1$$ from each side:

$$\frac{b^{2}}{a^{2}} = 3.$$

So we have the direct relation

$$b^{2} = 3a^{2}.$$

Next, because the given point $$(4,6)$$ lies on the hyperbola, it satisfies its equation. Hence we substitute $$x = 4$$ and $$y = 6$$ into

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

This gives

$$\frac{4^{2}}{a^{2}} - \frac{6^{2}}{b^{2}} = 1.$$

Simplifying the numerators, we write

$$\frac{16}{a^{2}} - \frac{36}{b^{2}} = 1.$$

We already know $$b^{2} = 3a^{2},$$ so we substitute it here:

$$\frac{16}{a^{2}} - \frac{36}{3a^{2}} = 1.$$

The second fraction simplifies because $$\frac{36}{3} = 12,$$ so we have

$$\frac{16}{a^{2}} - \frac{12}{a^{2}} = 1.$$

The numerators on the left combine to give

$$\frac{4}{a^{2}} = 1.$$

Now multiply both sides by $$a^{2}:$$

$$4 = a^{2}.$$

Hence we obtain

$$a^{2} = 4 \quad\text{and}\quad b^{2} = 3a^{2} = 12.$$

The complete equation of the hyperbola is therefore

$$\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1.$$

We are required to write the tangent at the point $$(x_1,y_1) = (4,6).$$ For a hyperbola of the form $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1,$$ the equation of the tangent at $$(x_{1},y_{1})$$ is given by the standard formula

$$\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 1.$$

Substituting $$x_{1}=4,\;y_{1}=6,\;a^{2}=4,\;b^{2}=12,$$ we have

$$\frac{x\cdot 4}{4} - \frac{y\cdot 6}{12} = 1.$$

Now each fraction simplifies:

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

To clear the denominator, multiply every term by $$2:$$

$$2x - y = 2.$$

Finally, we rewrite it in the form with all terms on the left-hand side:

$$2x - y - 2 = 0.$$

This exactly matches Option D.

Hence, the correct answer is Option D.