Question

The range of 'a'so that a².2 + 2xy + 4y2 = 0 represents distinct lines
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Answer 1

Step-by-step explanation:

If the equation 2 2 S ax hxy by gx fy c ≡ + + + + += 2 22 0 represents a pair of straight lines

then

i)

222 ∆≡ + − − − = abc fgh af bg ch 2 0 and (ii) 22 2 h ab g ac f bc ≥≥≥ , ,

Proof:

Let the equation S = 0 represent the two lines 1 11 lx my n + += 0 and 2 22 lx my n + += 0 .

Then

2 2

1 1 12 2 2

2 22

( )( ) 0

ax hxy by gx fy c

lx my n l x m y n

+ ++++

≡+ + + +=

Equating the co-efficients of like terms, we get

1 2 ll a = , 12 21 lm lm h + = 2 , m1m2 = b, and 12 21 l n ln + = 2g , 12 21 m n mn + = 2 f , 1 2 n n = c

(i) Consider the product(2 )(2 )(2 ) hg f

1 2 2 1 12 21 12 21 =+ + + ( )( )( ) l m l m ln l n mn m n

( ) ( )( ) 22 22 22 22 2 2 2 2

12 1 2 2 1 1 2 1 2 2 1 1 2 1 2 2 1 12 1 2 1 2 = + + ++ + + l l m n m n mm l n l n nn l m l m ll mm nn 2

2 2

12 1 2 2 1 1 2 1 2 1 2 1 2 2 1 12 1 2

2

1 2 1 2 2 1 12 1 2 12 1 2 1 2

[( ) 2 ] [( ) 2 ]

[( ) 2 ] 2

l l mn m n mm nn mm ln l n ll nn

nn lm l m ll mm ll mm nn

= + − + +−

+ +− +

2 22 = −+ −+ − a f bc b g ac c h ab (4 2 ) (4 2 ) (4 2 )

2 22 8 4[ ] fgh af bg ch ab = ++− c

2 22 ∴+ −− −= abc fgh af bg ch 2 0

ii)

2

2 12 21

12 1 2 2

lm l m h ab l l m m   + −= −    

2

1 2 2 1 12 1 2 ( )4

4

l m l m ll mm + −− =

2

12 21 ( ) 0

4

lm l m − = ≥

Similarly we can prove 2

g ≥ ac and 2 f b ≥ c

NOTE :

If 2 22 ∆= + − − − = abc fgh af bg ch 2 0 , 2 h ab ≥ , 2 g ac ≥ and 2 f bc ≥ , then the

equation 2 2 S ax hxy by gx fy c ≡ + + + + += 2 22 0 represents a pair of straight lines