Question 16

For (P): The radius of the circle is the perpendicular distance from $(1, 2)$ to $3x + 4y - 1 = 0$: $r = \frac{|3(1) + 4(2) - 1|}{\sqrt{3^2 + 4^2}} = \frac{10}{5} = 2$. Equation of circle: $(x - 1)^2 + (y - 2)^2 = 4$. Check points in List-II: Point $(3, 2)$ gives $(3 - 1)^2 + (2 - 2)^2 = 4$. Thus (P) $\rightarrow$ (3).

For (Q): For the parabola $y^2 = 8x$, $a = 2$. Tangent is $y = mx + \frac{2}{m}$. For it to touch $x^2 + y^2 = 2$, $\frac{|2/m|}{\sqrt{m^2 + 1}} = \sqrt{2} \Rightarrow \frac{2}{m^2(m^2 + 1)} = 1 \Rightarrow m^4 + m^2 - 2 = 0 \Rightarrow (m^2 + 2)(m^2 - 1) = 0$. Since $m > 0$, $m = 1$. Tangent is $y = x + 2$. Point $(7, 9)$ satisfies this. Thus (Q) $\rightarrow$ (2).

For (R): Ellipse is $\frac{x^2}{16} + \frac{y^2}{12} = 1$. $a = 4, b = \sqrt{12}, e = \sqrt{1 - 12/16} = 1/2$. End point of latus rectum $M = (ae, b^2/a) = (2, 3)$. Normal at $(2, 3)$ is $\frac{16x}{2} - \frac{12y}{3} = 16 - 12 \Rightarrow 8x - 4y = 4 \Rightarrow 2x - y = 1$. Point $(1, 1)$ satisfies this. Thus (R) $\rightarrow$ (1).

For (S): Hyperbola center $(0, 0)$, focus $ae = 5$, directrix $a/e = 16/5$. $(ae)(a/e) = 16 \Rightarrow a^2 = 16, e = 5/4$. $b^2 = a^2(e^2 - 1) = 16(25/16 - 1) = 9$. Equation is $\frac{x^2}{16} - \frac{y^2}{9} = 1$. Point $(8, 3\sqrt{3})$ satisfies $\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1$. Thus (S) $\rightarrow$ (5).

Matching: P-3, Q-2, R-1, S-5. Correct Option: B