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  • The locus of mid-points of the line segments joining and the points on the ellipse <img alt="\frac{x^{2}}{4}+\frac{y^{2}}{9}=1" src="https://entrancecorner.codecogs.co

The locus of mid-points of the line segments joining (-3,-5) and the points on the ellipse \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 is :
Option: 1 36 x^{2}+16 y^{2}+90 x+56 y+145=0
Option: 2 36 x^{2}+16 y^{2}+108 x+80 y+145=0
Option: 3 9 x^{2}+4 y^{2}+18 x+8 y+145=0
Option: 4 36 x^{2}+16 y^{2}+72 x+32 y+145=0

Answers (1)

best_answer

 

Let us take any parameteric point  B\left ( 2\cos\theta,3\sin\theta \right )  on the ellipse

Let  P_{A}\\ be the mid-point of  A\left ( -3,-5 \right ) \&\: B(h,k)\\

So   h=\frac{2\cos\theta-3}{2}\Rightarrow \cos\theta= \frac{2h+3}{2}\\

       k=\frac{3\sin\theta-5}{2}\Rightarrow \sin\theta=\frac{2k+5}{3}\\

Square and add;

\cos^{2}\theta+\sin^{2}\theta=\left ( \frac{2h+3}{2} \right )^{2}+\left ( \frac{2k+5}{3} \right )^{2}=1\\

Replace  h\rightarrow x\&\; k\rightarrow y\\

\Rightarrow \frac{4x^{2}+12x+9}{4}+\frac{4y^{2}+20y+25}{9}=1\\

\Rightarrow 36x^{2}+108x+81+16y^{2}+80y+100-36\\

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

Posted by

Kuldeep Maurya

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