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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 :
$9 x^{2}+4 y^{2}+18 x+8 y+145=0$
$36 x^{2}+16 y^{2}+90 x+56 y+145=0$
$36 x^{2}+16 y^{2}+108 x+80 y+145=0$
$36 x^{2}+16 y^{2}+72 x+32 y+145=0$
Solution
General point on $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is $\mathrm{A}(2 \cos \theta, 3 \sin \theta)$
given $\mathrm{B}(-3,-5)$
$\text { midpoint } C\left(\frac{2 \cos \theta-3}{2}, \frac{3 \sin \theta-5}{2}\right)$
$\mathrm{h}=\frac{2 \cos \theta-3}{2} ; \mathrm{k}=\frac{3 \sin \theta-5}{2}$
$\Rightarrow\left(\frac{2 \mathrm{~h}+3}{2}\right)^{2}+\left(\frac{2 \mathrm{k}+5}{3}\right)^{2}=1$
$\Rightarrow 36 \mathrm{x}^{2}+16 \mathrm{y}^{2}+108 \mathrm{x}+80 \mathrm{y}+145=0$
