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JEE Main 202127 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
27 August
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Straight Lines

Let A be a fixed point ( 0, 6 ) and B be a moving point ( 2t, 0 ) ....

Let AA be a fixed point (0,6)( 0, 6 ) and BB be a moving point (2t,0).( 2t, 0 ) . Let MM be the mid-point of ABAB and the perpendicular bisector of ABAB meets the y−y -axis at C.C . The locus of the mid-point PP of MCMC is

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

Options

  1. A3x2+2y−6=03{x}^{2} + 2y - 6 = 0
  2. B2x2−3y+9=02{x}^{2} - 3y + 9 = 0
  3. C3x2−2y−6=03{x}^{2} - 2y - 6 = 0
  4. D2x2+3y−9=02{x}^{2} + 3y - 9 = 0

Correct answer

D. 2x2+3y−9=02{x}^{2} + 3y - 9 = 0

Step-by-step solution

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

Mid point M=(t,3)M = \left(t, 3\right)

mAB=−3t{m}_{AB} = \dfrac{- 3}{t}

Perpendicular bisector of ABAB is

y−3=t3(x−t)y - 3 = \dfrac{t}{3}\left(x - t\right)

So, C=(0,3−t23)C = \left(0, 3 - \dfrac{{t}^{2}}{3}\right)

Let PP be \left(h, k\right)</p><p></p><p>h = \dfrac{t}{2}; k = 3 - \dfrac{{t}^{2}}{6}</p><p></p><p>⇒ k = 3 - \dfrac{4{h}^{2}}{6}</p><p></p><p>⇒ 2{x}^{2} + 3y - 9 = 0

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