Let A be a fixed point (0,6) and B be a moving point (2t,0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
(A)2x2+3y−9=0✓ Correct
(B)3x2+2y−6=0
(C)2x2−3y+9=0
(D)3x2−2y−6=0
Answer: (A)
Solution
With A(0,6) and B(2t,0) the midpoint is M(t,3), and AB has slope −t3, so the perpendicular bisector through M has slope 3t: y−3=3t(x−t). Setting x=0 gives C(0,3−3t2). The midpoint P(h,k) of MC satisfies h=2t and k=3−6t2. Eliminating t=2h, k=3−32h2, so 3k=9−2h2. The locus is 2x2+3y−9=0.
JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option