Match the following :
Column-I | Column-II |
(i) The number of complex numbers satisfying |z + 2i| + |z – 2i| = 8 and |z – i| + |z + i| = 2 is | [A] 5 |
(ii) The value of 3 + (cot14º – 1) (cot 31º – 1) is | [B] 0 |
(iii) Any chord of the conic x2 + y2 + xy = 1 through (0, 0) is bisected at (p, q) then p + q is equal to | [C] 9 |
(iv) Line L1 : 3x + 4y + 12 = 0 is rotated by an angle of tan–1 in anticlockwise direction with respect to the point, where it cuts the x-axis. If line obtained in new position is shown by L2 = 0. If the incentre of the triangle Formed by L1 = 0, L2 = 0 and y-axis is (4λ, 3k) then 10k – 24λ is equal to | [D] 4 |
Text Solution
Verified by Experts(i) [B]; (ii) [A]; (iii) [B]; (iv) [C]
Ans.
(i) [B]
(ii) [A]
(iii) [B]
(iv) [C]
Sol. (i) First equation represents an ellipse and second
equation represents a line segment joining (0, 1) and (0, –1) completely contained inside the ellipse. So number of solution is zero.
(ii) 3 + (cot14º – 1)(cot31º – 1)
= 3 + 
= 3 + 
= 3 + 
= 3 + 2 = 5
[ 1 – tan14º – tan31º = tan14º tan31º]
(iii) As the given curve is an ellipse
( Δ ≠ 0 and h 2 < ab) (0, 0) is the centre of the given
ellipse. Any chord passing through centre is bisected by it.
∴ (p, q) = (0, 0) ⇒ p + q = 0
(iv) θ = tan –1 
= 
m = 
∴ equation of the line L 2 = 0 is
L 2 : 3x – 4y + 12 = 0
Now incentre of the Δ ABC is
≡ 
≡ (4 λ , 3k)
∴ λ = 
, k = 0
⇒ 10k – 24 λ = 9
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