Let the given tangent be $\sqrt{3}x+2y=4.$ First rewrite it in slope form: $2y=-\sqrt{3}x+4 \implies y=-\frac{\sqrt{3}}{2}x+2.$ So the slope is $m=-\frac{\sqrt{3}}{2}.$ The given point of contact is $\left(\sqrt{3},\frac12\right).$

We now match this with the entries in the columns.

1. Identify the conic from Column 3 (point of contact)

We test which point-of-contact formula can produce $\left(\sqrt{3},\frac12\right).$

Case (R):

$\left(\frac{-a^2m}{\sqrt{a^2m^2+1}},\frac{1}{\sqrt{a^2m^2+1}}\right)$ Using $m=-\frac{\sqrt3}{2}$ and equating the second coordinate to $\frac12$: $\frac{1}{\sqrt{a^2m^2+1}}=\frac12$ $\sqrt{a^2m^2+1}=2$ $a^2m^2=3.$ Since $m^2=\frac34,$ we get $a^2\cdot \frac34=3 \implies a^2=4 \implies a=2.$ Now check the first coordinate: \frac{-a^2m}{\sqrt{a^2m^2+1}}=rac{-4\left(-\frac{\sqrt3}{2}\right)}{2}=\sqrt3, which matches.

So the point corresponds to (R).

2. Identify the tangent equation from Column 2

Now test with $m=-\frac{\sqrt3}{2}$ and $a=2$.

Equation (iv):

$y=mx+a\sqrt{m^2+1}$ Substitute values: $y=-\frac{\sqrt3}{2}x+2\sqrt{\frac34+1}$ $= -\frac{\sqrt3}{2}x+2\sqrt{\frac74}$ $= -\frac{\sqrt3}{2}x+\sqrt7,$ not our line.

Equation (ii):

$my=m^2x+a$ Substitute $m=-\frac{\sqrt3}{2}$ and $a=2$: $-\frac{\sqrt3}{2}y=\frac34x+2.$ This is not the given line.

Equation (iii):

$y=mx+\sqrt{a^2m^2-1}$ Substitute $a=2$: $y=-\frac{\sqrt3}{2}x+\sqrt{4\cdot\frac34-1}$ $= -\frac{\sqrt3}{2}x+\sqrt{3-1}$ $= -\frac{\sqrt3}{2}x+\sqrt2,$ not our line.

Equation (iv) of hyperbola-type entry in standard matching

For conic $x^2-a^2y^2=a^2$, tangent of slope $m$ is $y=mx+\sqrt{a^2m^2-1}.$ Now with $a=2$ and $m=-\frac{\sqrt3}{2}$: $y=-\frac{\sqrt3}{2}x+\sqrt{4\cdot\frac34-1}$ $= -\frac{\sqrt3}{2}x+\sqrt2,$ again not the given line.

So instead let us identify directly from the conic using the point.

3. Determine which conic passes through the point and has the given tangent

Option labels suggest checking Column 1 entries.

Conic (II):

This is the ellipse $x^2+a^2y^2=a^2.$ At the point $\left(\sqrt3,\frac12\right)$, $3+a^2\cdot\frac14=a^2$ $3=\frac34a^2$ $a^2=4 \implies a=2.$ So the ellipse is $x^2+4y^2=4.$ Tangent to $x^2+4y^2=4$ at $(x_1,y_1)$ is $xx_1+4yy_1=4.$ At $\left(\sqrt3,\frac12\right)$: $\sqrt3 x+4y\cdot\frac12=4$ $\sqrt3 x+2y=4,$ which matches exactly.

Hence the conic is (II).

4. Match the tangent formula and point formula for this conic

For ellipse $x^2+a^2y^2=a^2,$ its tangent of slope $m$ is $y=mx+\sqrt{a^2m^2+1},$ which is entry (iv) in Column 2.

Its point of contact corresponding to slope-form tangent is $\left(\frac{-a^2m}{\sqrt{a^2m^2+1}},\frac{1}{\sqrt{a^2m^2+1}}\right),$ which is (R) in Column 3.

Thus the correct combination is $(II),(iv),(R).$

5. Final answer

Therefore the only correct option is $\boxed{\text{B}}.$