Q: The locus of the middle points of chords of hyperbola 3x2 – 2y2 + 4x – 6y = 0 parallel to y = 2x is
(A) 3x – 4y = 4
(B) 3y – 4x + 4 = 0
(C) 4x – 4y = 3
(D) 3x – 4y = 2
Ans: (A)
Sol. Let mid point be (h, k).
Equation of a chord whose mid point is (h, k) would be T = S1
or 3xh – 2yk + 2( x+h) – 3( y+k) = 3h2 – 2k2 + 4h – 6k
⇒ x(3k + 2) – y ( 2k+3) – 2h + 3k – 3h2 + 2k2 = 0
it’s slope is $\large \frac{3h+2}{2k+3} = 2 $(given)
⇒ 3h = 4k + 4
⇒ Required locus is 3x – 4y = 4.
Hence (A) is the correct answer.