Question
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 3 )
Step-by-step solution
For (P): The radius of the circle is the perpendicular distance from (1, 2) to 3x + 4y - 1 = 0. r = |3(1) + 4(2) - 1| 3^2 + 4^2 = 10 5 = 2 The equation of the circle is (x - 1)^2 + (y - 2)^2 = 4. Checking the given points, (3, 2) satisfies the equation. Thus, (P) (3). For (Q): The equation of a tangent to y^2 = 8x with slope m is y = mx + 2 m . Since it is also a tangent to x^2 + y^2 = 2, the perpendicular distance from (0, 0) to the line mx - y + 2 m = 0 is equal to the radius 2 . | 2 m | m^2 + 1 = 2 4 m^2 = 2(m^2 + 1) m^4 + m^2 - 2 = 0 (m^2 + 2)(m^2 - 1) = 0 Since m is real and positive, m = 1. The equation of the common tangent is y = x + 2. Checking the given points, (7, 9) satisfies the equation. Thus, (Q) (2). For (R): The equation of the ellipse is x^2 16 + y^2 12 = 1. Here a^2 = 16 and b^2 = 12. Eccentricity e = 1 - 12 16 = 1 2 . The end point of the latus rectum in the first quadrant is (ae, b^2 a ) = (4 1 2 , 12 4 ) = (2, 3). The equation of the normal at (2, 3) is 16x 2 - 12y 3 = 16 - 12. 8x - 4y = 4 2x - y = 1. Checking the given points, (1, 1) satisfies the equation. Thus, (R) (1). For (S): The centre is (0, 0). The focus is (ae, 0) = (5, 0) ae = 5. The directrix is x = - 16 5 a e = 16 5 . Multiplying the two equations gives a^2 = 16. Dividing the two equations gives e^2 = 25 16 . b^2 = a^2(e^2 - 1) = 16 ( 25 16 - 1 ) = 9. The equation of the hyperbola is x^2 16 - y^2 9 = 1. Checking the given points, (8, 3 3 ) satisfies the equation. Thus, (S) (5). Answer: (P) (3), (Q) (2), (R) (1), (S) (5)