JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 6)
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 :
$$9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$$
$$36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$$
$$36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$
$$36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$$
Explanation
General point on $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ is A(2cos$$\theta$$, 3sin$$\theta$$)
given B($$-$$3, $$-$$5)
midpoint $$C\left( {{{2\cos \theta - 3} \over 2},{{3\sin \theta - 5} \over 2}} \right)$$
$$h = {{2\cos \theta - 3} \over 2};k = {{3\sin \theta - 5} \over 2}$$
$$ \Rightarrow {\left( {{{2h + 3} \over 2}} \right)^2} + {\left( {{{2k + 5} \over 3}} \right)^2} = 1$$
$$ \Rightarrow 36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$$
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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