JEE Main · 2009 · 12 Apr 2009 · Offline · Ellipse

The inner ellipse $x^2+4y^2=4$ has semi-axes 2 and 1, so the rectangle snug around it has corners at $(\pm 2,\pm 1)$. The outer ellipse $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ passes through $(4,0)$, giving $A^2=16$, and it must also pass through the corner $(2,1)$, so $\frac{4}{16}+\frac{1}{B^2}=1$ leads to $B^2=\frac{4}{3}$. The outer ellipse is thus $\frac{x^2}{16}+\frac{3y^2}{4}=1$, that is $x^2+12y^2=16$.