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Question : 129 of 160
Marks:
+1 ,
-0
Solution:
Given that, equation of pair of straight lines is
( a x + b y ) 2 − 3 ( b x − a y ) 2 = 0 ‌ ⇒ a 2 x 2 + b 2 y 2 + 2 a b x y ‌ ‌ − 3 ( b 2 x 2 + a 2 y 2 − 2 a b x y ) = 0 ‌ ⇒ ‌ ‌ ( a 2 − 3 b 2 ) x 2 − ( b 2 − 3 a 2 ) y 2 + 8 a b x y = 0 Let the equation of line be
y = m 1 x , y = m 2 x .
∴ ‌ ‌ m 1 + m 2 = ‌ ‌ ‌ [ ] ‌ ‌ and ‌ ‌ m 1 m 2 = ‌ ‌ ‌ Now, ‌ ( m 1 − m 2 ) 2 = ( m 1 + m 2 ) 2 − 4 m 1 m 2 ‌ ‌ = ‌ − ‌ ‌ ‌ = ‌ ‌ ‌ [ 4 a 2 b 2 − ( a 2 − 3 b 2 ) ( b 2 − 3 a 2 ) ] ‌ ‌ = ‌ [ 4 a 2 b 2 − ( a 2 b 2 − 3 a 4 − 3 b 4 + 9 a 2 b 2 ) ] ‌ = ‌ [ − 6 a 2 b 2 + 3 a 4 + 3 b 4 ] ‌ = ‌ ( a 2 − b 2 ) 2 ∴ ‌ ‌ ( m 1 − m 2 ) = ‌ ⋅ ( a 2 − b 2 ) The intersecting point on the lines
y = m 1 x ,
y = m 2 x ‌ and ‌ a x + b y + c = 0 are
A ( 0 , 0 ) , B ( ‌ , ‌ ) ‌ ‌ and ‌ C ( ‌ , ‌ ) ‌ ‌ Area of ‌ ∆ = ‌ | | ‌ = ‌ [ ‌ − ‌ ] ‌ = − ‌ [ ‌ c 2 ( m 1 − m 2 ) a 2 + a b ( m 1 + m 2 ) + b 2 m 1 m 2
] ‌ = ‌ − [ ⋅ ( a 2 − b 2 ) ] a 2 + a b ( ‌ ) + b 2 ( ‌ )
‌ [from Eq. (i)] ‌ ‌ ‌ = ‌ − √ 3 [ ( a 2 − b 2 ) c 2 ] a 2 ( b 2 − 3 a 2 ) − 4 a 2 b 2 + a 2 b 2 − 3 b 4
‌ = ‌ − √ 3 [ c 2 ( a 2 − b 2 ) ] a 2 b 2 − 3 a 4 − 3 a 2 b 2 − 3 b 4
‌ = ‌ − √ 3 c 2 − 3 ( a 4 + b 4 + 2 a 2 b 2 )
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