JEE Main 2021 — Mathematics Coordinate Geometry

Parametric point on the given ellipse is $(2\mathrm{sin}\theta ,3\mathrm{cos}\theta )$(2sinθ,3cosθ)

Let, the mid-point of line segments joining $(-3,-5)$(−3,−5) and $(2\mathrm{sin}\theta ,3\mathrm{cos}\theta )$(2sinθ,3cosθ) is $(h,k)$(h,k)

Then, by using mid-point formula, we get

$$\frac{2\mathrm{sin}\theta -3}{2}=h,\frac{3\mathrm{cos}\theta -5}{2}=k$$22sinθ−3​=h,23cosθ−5​=k

$$\Rightarrow 2\mathrm{sin}\theta =2h+3,3\mathrm{cos}\theta =2k+5$$⇒2sinθ=2h+3,3cosθ=2k+5

$$\Rightarrow \mathrm{sin}\theta =\frac{2h+3}{2},\mathrm{cos}\theta =\frac{2k+5}{3}$$⇒sinθ=22h+3​,cosθ=32k+5​

We know, ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$sin2θ+cos2θ=1

$$\Rightarrow {(\frac{2h+3}{2})}^{2}+{(\frac{2k+5}{3})}^{2}=1$$⇒(22h+3​)2+(32k+5​)2=1

$$\Rightarrow \frac{1}{4}[4{h}^{2}+9+12h]+\frac{1}{9}[4{k}^{2}+25+20k]=1$$⇒41​[4h2+9+12h]+91​[4k2+25+20k]=1

$$\Rightarrow 36{h}^{2}+16{k}^{2}+108h+80k+145=0$$⇒36h2+16k2+108h+80k+145=0

Hence, the locus of $(x,y)$(x,y) is $36{x}^{2}+16{y}^{2}+108x+80y+145=0$36x2+16y2+108x+80y+145=0