Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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

  1. A $9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$
  2. B $36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$
  3. C $36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$ Correct answer
  4. D $36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$

Solution

General point on ${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$ is A(2cos$\theta$, 3sin$\theta$)<br><br>given B($-$3, $-$5)<br><br>midpoint $C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)$<br><br>$h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}$<br><br>$$ \Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1$$<br><br>$\Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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