Question 16

MATRIX MATCHHARD

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

List - I

The circle with centre $(1, 2)$ and touching the straight line $3x + 4y = 1$ , passes through

The common tangent to the circle $x^2 + y^2 = 2$ and the parabola $y^2 = 8x$ with positive slope, passes through

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

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

List-II

the point $(1, 1)$

the point $(7, 9)$

the point $(3, 2)$

the point $(2, 5)$

the point $(8, 3\sqrt{3})$

(A)

P-3, Q-4, R-1, S-2

(B)

P-3, Q-2, R-1, S-5

(C)

P-3, Q-2, R-4, S-5

(D)

P-4, Q-1, R-2, S-3

Detailed Solution

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

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