Hyperbola

Concept:Matching problems require solving each geometric condition to find the correct point from List-II.Explanation:(P) Radius $r = \frac{|3(1)+4(2)-1|}{\sqrt{3^2+4^2}} = \frac{10}{5}=2$.Circle: $(x-1)^2+(y-2)^2=4$.Check $(3,2)$: $(2)^2+(0)^2=4$, so passes through $(3,2)$ → P→3.(Q) Tangent to parabola $y^2=8x$: $y=mx+\frac{2}{m}$ (since $a=2$).Circle $x^2+y^2=2$: radius $\sqrt{2}$, centre $(0,0)$.Distance from centre to line $mx-y+\frac{2}{m}=0$ equals $\sqrt{2}$: $\frac{|2/m|}{\sqrt{m^2+1}}=\sqrt{2}$.Squaring: $\frac{4}{m^2}=2(m^2+1)\Rightarrow m^4+m^2-2=0\Rightarrow m^2=1$.Positive slope: $m=1$, so line $y=x+2$.Check $(7,9)$: $9=7+2$ → Q→2.(R) Ellipse $3x^2+4y^2=48\Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1$, $a^2=16$, $b^2=12$, $e=\sqrt{1-\frac{12}{16}}=\frac12$.Latus rectum point $M$ (first quadrant): $x=ae=2$, $y=\frac{b^2}{a}=3$ → $M(2,3)$.Normal at $M$: $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2 e^2\Rightarrow \frac{16x}{2}-\frac{12y}{3}=4\Rightarrow 8x-4y=4\Rightarrow 2x-y=1$.Check $(1,1)$: $2-1=1$ → R→1.(S) Hyperbola centre $(0,0)$, focus $(5,0)\Rightarrow ae=5$. Directrix $5x+16=0\Rightarrow x=-\frac{16}{5}\Rightarrow \frac{a}{e}=\frac{16}{5}$.Multiply: $a^2=5\cdot\frac{16}{5}=16\Rightarrow a=4$, $e=\frac{5}{4}$.$b^2=a^2(e^2-1)=16(\frac{25}{16}-1)=9$. Equation: $\frac{x^2}{16}-\frac{y^2}{9}=1$.Check $(8,3\sqrt{3})$: $\frac{64}{16}-\frac{27}{9}=4-3=1$ → S→5.Thus (P)→3, (Q)→2, (R)→1, (S)→5.Answer:Option (B) is correct.