Question

6. The pair of lines √3x²- 4xy + √3y² = 0 are rotated
about the origin by π/6 in the anticlockwise sense. The
equation of the pair of lines in the new position is
(a) √3x² - xy = 0
(b) √3x² + xy = 0
(c) xy - √3x = 0


don't write any unnecessary answer otherwise I will report​

Answer 1

Answer:

C is the answer.,,,,,,

Answer 2

Solution :- Given pair of lines √3x²-4xy + √3y² = 0

=> √3x² - 3xy-xy +√3y² = 0

=> √3x(x-√3y) - y(x-√3y) =0

=> (√3x-y) (x-√3y) =0 rewritten form

(√3x-y) = 0

y = √3x, But we know √3 = tan60°

y = tan60° x

According to given in position ,the pair of lines rotate about π/6 = 30°

∴y = tan90° x

y = 1/0 x , we know that tan90° = ∞ = 1/0

x = 0______(1)

And, (x-√3y) = 0

y = x/√3 , we know. 1/√3 = tan30°

y = tan30° x

∴y = tan60° x

y = √3x

y - √3x _______(2)

Multiplying equation (1) and (2)

x(y-√3x) = 0

xy - √3x²= 0

or √3x²-xy = 0 option (A) is correct !

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