The locus of the middle points of chords of hyperbola 3x2−2y2+4x−6y=0 parallel to y = 2x, is
A
3x – 4y = 4
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B
3y – 4x + 4 = 0
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C
4x – 4y = 3
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D
3x – 4y = 2
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Solution
The correct option is A 3x – 4y = 4 By T = S1, the equation of chord whose mid-point is (α,β) is 3xα−2yβ+(x+α)−2(y+β)=0 ⇒x(3α+2)−y(2β+3)=2 as it is parallel to y = 2x ∴3α–4β=4 ∴ Required locus is 3x – 4y = 4