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 $(8, 3\sqrt{3})$ |

$(\text{P})\ \text{circle centre } (1,2) \perp\!\!\text{-tangent to } 3x+4y=1$ $(\text{Q})\ \text{common tangent (slope} > 0\text{) of } x^{2}+y^{2}=2,\ y^{2}=8x$ $(\text{R})\ \text{normal at latus-rectum end M of ellipse } 3x^{2}+4y^{2}=48$ $(\text{S})\ \text{hyperbola H: focus } (5,0),\ \text{directrix } 5x+16=0$ $\text{To find: the correct matching option (1)--(4)}$ $\text{Plan: solve each of } (P), (Q), (R), (S)$ $\text{then match with a point in List-II, then pick the option}$ $r = \dfrac{|3(1) + 4(2) - 1|}{\sqrt{3^{2}+4^{2}}} = \dfrac{10}{5} = 2$ $(x-1)^{2} + (y-2)^{2} = 4$ $(3,2):\ (3-1)^{2} + (2-2)^{2} = 4 \ \checkmark$ $(\text{P}) \to (3)$ $\text{Common tangents: } y = x + 2,\quad y = -x - 2$ $\text{Positive slope} \Rightarrow y = x + 2$ $(7,9):\ 9 = 7 + 2 \ \checkmark$ $(\text{Q}) \to (2)$ $3x^{2} + 4y^{2} = 48 \Rightarrow \dfrac{x^{2}}{16} + \dfrac{y^{2}}{12} = 1$ $e = \sqrt{1 - \dfrac{12}{16}} = \dfrac{1}{2},\quad \text{latus rectum: } x = \pm ae = \pm 2$ $M = \left(2,\ \dfrac{b^{2}}{a}\right) = (2, 3)$ $\text{Normal at } M:\ y - 3 = 2(x-2) \Rightarrow y = 2x - 1$ $(1,1):\ 1 = 2(1) - 1 \ \checkmark$ $(\text{R}) \to (1)$ $\text{Focus } (ae,0) = (5,0) \Rightarrow ae = 5$ $\text{Directrix } x = -\dfrac{16}{5} \Rightarrow \dfrac{a}{e} = \dfrac{16}{5}$ $a^{2} = 16 \Rightarrow a = 4,\quad e = \dfrac{5}{4},\quad b^{2} = a^{2}(e^{2}-1) = 9$ $\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1$ $(8, 3\sqrt{3}):\ \dfrac{64}{16} - \dfrac{27}{9} = 4 - 3 = 1 \ \checkmark$ $(\text{S}) \to (5)$ $(\text{P})\to(3),\ (\text{Q})\to(2),\ (\text{R})\to(1),\ (\text{S})\to(5) \Rightarrow \text{option (2)}$