By appropriately matching the information given in the three columns of the following table.

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

	Column - 1	Column - 2	Column - 3
(i)	${x^2} + {y^2} = a$	$my = {m^2}x + a$	$$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$
(ii)	${x^2}{a^2}{y^2} = {a^2}]$	$y = mx + a\sqrt {{m^2} + 1}$	$$\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},\,{a \over {\sqrt {{m^2} + 1} }}} \right)$$
(iii)	${y^2} = 4ax$	$y = mx + \sqrt {{a^2}{m^2} - 1}$	$$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},\,{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$$
(iv)	${x^2} - {a^2}{y^2} = {a^2}$	$y = mx + \sqrt {{a^2}{m^2} + 1}$	$$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$

The tangent to a suitable conic (Column 1) at $$\left( {\sqrt 3 ,\,{1 \over 2}} \right)$$ is found to be $\sqrt 3 x + 2y = 4$, then which of the following options is the only CORRECT combination?