JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Ellipse

Let $P(x,y)$ be the moving point. With $O(0,0)$ and $A(0,1)$ fixed and $OA=1$, a perimeter of $4$ means $OP+AP=3$, the defining condition of an ellipse with foci $O$ and $A$. Writing it out, $\sqrt{x^2+y^2}+\sqrt{x^2+(y-1{)}^2}=3$; isolating one radical and squaring gives $y+4=3\sqrt{x^2+y^2}$, and squaring again gives $y^2+8y+16=9x^2+9y^2$, which rearranges to $9x^2+8y^2-8y=16$.