Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, where a > b > 0, be a hyperbola in the XY-plane whose conjugate axis LM subtends an angle of 60$$^\circ $$ at one of its vertices N. Let the area of the $$\Delta $$LMN be $$4\sqrt 3 $$.

	List - I		List - II
P.	The length of the conjugate axis of H is	1.	8
Q.	The eccentricity of H is	2.	$${4 \over {\sqrt 3 }}$$
R.	The distance between the foci of H is	3.	$${2 \over {\sqrt 3 }}$$
S.	The length of the latus rectum of H is	4.	4

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?

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