Let S be the circle in the XY-plane defined the equation x² + y² = 4.

Let E₁E₂ and F₁F₂ be the chords of S passing through the point P₀ (1, 1) and parallel to the X-axis and the Y-axis, respectively. Let G₁G₂ be the chord of S passing through P₀ and having slope$$-$$1. Let the tangents to S at E₁ and E₂ meet at E₃, then tangents to S at F₁ and F₂ meet at F₃, and the tangents to S at G₁ and G₂ meet at G₃. Then, the points E₃, F₃ and G₃ lie on the curve

Let S be the circle in the XY-plane defined the equation x² + y² = 4.

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

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)	$${x^2} - {a^2}{y^2} = {a^2}$$	$$y = mx + \sqrt {{a^2}{m^2} + 1} $$	$$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$

For $$a = \sqrt 2 $$, if a tangent is drawn to a suitable conic (Column 1) at the point of contact ($$-$$1, 1), then which of the following options is the only CORRECT combination for obtaining its equation?

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?