The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$
So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$
If $$PQRS$$ is the rectangular in which it is inscribed, then
$$P = \left( {2,1} \right).$$
Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
be the ellipse circumscribing the rectangular $$PQRS$$.
Then it passes through $$P\,\,(2,1)$$
$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$
Also, given that, it passes through $$(4,0)$$
$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$
$$ \Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$
$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$
or $${x^2} + 12y{}^2 = 16$$