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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 :

JEE · Math · previous-year question

  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
  4. D.$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$

Answer

C. $$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$

Explanation

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

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