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?

Let $$P(6, 3)$$ be a point on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal at the point $$P$$ intersects the $$x$$-axis at $$(9, 0)$$, then the eccentricity of the hyperbola is

The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of the circle with $$AB$$ as its diameter is

The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is