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QJEEM21Ma298f: JEE Main 2021 Mathematics - Class 11 JEEM21Ma298f Solution: The locus of mid-points of the line segments joining ( 3, 5)…

The locus of mid-points of the line segments joining ( −- 3, −- 5) and the points on the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1 is :

Multiple Choice Options

  • A 9x2+4y2+18x+8y+145=09{x^2} + 4{y^2} + 18x + 8y + 145 = 0
  • B 36x2+16y2+90x+56y+145=036{x^2} + 16{y^2} + 90x + 56y + 145 = 0
  • C 36x2+16y2+108x+80y+145=036{x^2} + 16{y^2} + 108x + 80y + 145 = 0
  • D 36x2+16y2+72x+32y+145=036{x^2} + 16{y^2} + 72x + 32y + 145 = 0

Step-by-Step Verified Solution

General point on x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1 is A(2cos θ\theta , 3sin θ\theta )

given B( −- 3, −- 5)

midpoint C(2cos⁡θ−32,3sin⁡θ−52)C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)

h=2cos⁡θ−32;k=3sin⁡θ−52h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}

⇒(2h+32)2+(2k+53)2=1\Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1

⇒36x2+16y2+108x+80y+145=0\Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0

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