A hyperbola, having the transverse axis of length $$2\sin \theta ,$$ is confocal with the ellipse $$3{x^2} + 4{y^2} = 12.$$ Then its equation is

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I	List-II
(P) The circle with centre $(1,2)$ and touching the straight line $$3x + 4y = 1$$ passes through	(1) the point $(1,1)$
(Q) The common tangent to the circle $$x^2 + y^2 = 2$$ and the parabola $$y^2 = 8x$$ with positive slope, passes through	(2) the point $(7,9)$
(R) Let $M$ be the end point of the latus rectum of the ellipse $$3x^2 + 4y^2 = 48$$ such that $M$ lies in the first quadrant. Then the normal to the ellipse drawn at $M$ passes through	(3) the point $(3,2)$
(S) Let $H$ be the hyperbola whose centre is at the origin, one of the foci is at $(5,0)$, and one directrix is $$5x + 16 = 0$$ Then $H$ passes through	(4) the point $(2,5)$
	(5) the point $(8, 3\sqrt{3})$

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?