1. Write the given ellipse in standard form

The ellipse is $x^2+4y^2=4$ Dividing by $4$, $\frac{x^2}{4}+\frac{y^2}{1}=1$ So its semi-axes are:

• along $x$-axis: $a=2$
• along $y$-axis: $b=1$

Hence the rectangle aligned with the coordinate axes and circumscribing this ellipse has vertices $(\pm 2,\pm 1).$

2. Now this rectangle is inscribed in another ellipse

Since the outer ellipse is also aligned with the coordinate axes, let its equation be $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ with semi-axes $A,B$.

Because the rectangle is inscribed in this ellipse, each vertex of the rectangle lies on it. So point $(2,1)$ lies on the outer ellipse: $\frac{2^2}{A^2}+\frac{1^2}{B^2}=1$ $\frac{4}{A^2}+\frac{1}{B^2}=1 \quad ...(1)$

3. Use the condition that the outer ellipse passes through $(4,0)$

Substitute $(4,0)$ into $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ we get $\frac{16}{A^2}=1$ So, $A^2=16.$

4. Find $B^2$ using equation (1)

Substitute $A^2=16$ into (1): $\frac{4}{16}+\frac{1}{B^2}=1$ $\frac14+\frac{1}{B^2}=1$ $\frac{1}{B^2}=\frac34$ $B^2=\frac43.$

Thus the outer ellipse is $\frac{x^2}{16}+\frac{y^2}{4/3}=1.$

5. Convert to standard polynomial form

Multiply by $16$: $x^2+12y^2=16.$

6. Check options

• A: $x^2+12y^2=16$ ✅
• B: $4x^2+48y^2=48 \Rightarrow x^2+12y^2=12$ ❌
• C: $4x^2+64y^2=48 \Rightarrow x^2+16y^2=12$ ❌
• D: $x^2+16y^2=16$ ❌

Therefore, the correct option is A.