$$\begin{gathered} \text{If the line }\alpha x+2y-4=\mathrm{0 }\text{is the tangent to the ellipse}\mathrm{ 3}{x}^2+4y^2=\mathrm{1 }\text{then}\mathrm{ \alpha }\text{is:} \\ \end{gathered}$$

The correct answer is 3\sqrt{5}.

Step 1: Express the line in slope-intercept form
The given equation of the line is:

\alpha x+2y-4=0

Rearranging this into slope-intercept form (y=mx+c):

2y=-\alpha x+4

y=\left(-\frac{\alpha }{2}\right)x+2

Comparing this with y=mx+c, we identify:

m=-\frac{\alpha }{2}

c=2

Step 2: Express the ellipse in standard form
The given equation of the ellipse is:

3x^2+4y^2=1

Rewriting this in the standard form of an ellipse, \frac{x^2}{a^2}+\frac{y^2}{b^2}=1:

\frac{x^2}{1/3}+\frac{y^2}{}=1

Comparing with the standard equation, we get:

a^2=\frac{1}{3}

b^2=\frac{1}{4}

Step 3: Apply the condition of tangency
For a line y=mx+c to be tangent to an ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, the condition of tangency is:

c^2=a^2{m}^2+b^2

Substitute the known values into the tangency condition:

2^2=\left(\frac{1}{3}\right){\left(-\frac{\alpha }{2}\right)}^2+\frac{1}{4}

Step 4: Solve for \alpha

4=\frac{1}{3}\cdot \frac{{\alpha }^2}{4}+\frac{1}{4}

4=\frac{{\alpha }^2}{12}+\frac{1}{4}

Subtract \frac{1}{4} from both sides:

4-\frac{1}{4}=\frac{{\alpha }^2}{12}

\frac{15}{4}=\frac{{\alpha }^2}{12}

Multiply both sides by 12:

{\alpha }^2=\frac{15}{4}\times 12

{\alpha }^2=15\times 3=45

Taking the positive square root to match the provided option:

\alpha =\sqrt{45}=\sqrt{9\times 5}=3\sqrt{5}