Question
Question: If a and b are positive numbers (a \< b), then the range of values of K for which a real l can be fo...
If a and b are positive numbers (a < b), then the range of values of K for which a real l can be found such that the equation ax2 + 2lxy + by2 + 2K(x + y + 1) = 0 represents a pair of straight lines is –
A
a< K2< b
B
a £ K2£ b
C
K2£ a or K2³ b
D
K £ 2a or K ³ 2b
Answer
K £ 2a or K ³ 2b
Explanation
Solution
We have ax2 + 2lxy + by2 + 2Kx + 2Ky + 2K = 0
h = l, g = K, c = 2K, f = K
= abc + 2fgh – af2 – bg2 – ch2 = 0
ab.(2K) + 2lK2 – aK2 – bK2 – 2l2K = 0
2Kl2 – 2K2l + (a + b) K2 – 2abK = 0
For real l, B2 – 4AC ³ 0
4K4 – 4 · 2K [(a + b) K2 – 2abK] ³ 0
K2 – 2(a + b) K + 4ab ³ 0, (K – 2a) (K – 2b) ³ 0
K £ 2a or K ³ 2b.