The ellipse $$x^2 + 4y^2 = 4$$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $$(4, 0)$$. Then the equation of the ellipse is

The inner ellipse is $$x^{2}+4y^{2}=4$$. Compare with $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ to get $$a=2,\;b=1$$. Hence the extreme points of this ellipse are $$\pm(2,0)$$ and $$\pm(0,1)$$.

Since the ellipse is “inscribed in a rectangle aligned with the coordinate axes”, the smallest such rectangle has sides $$x=\pm 2$$ and $$y=\pm 1$$. Therefore its four vertices are $$(\pm 2,\pm 1)$$.

This rectangle is now “inscribed in another ellipse”. Let that outer ellipse have its axes along the coordinate axes, so its equation is assumed to be $$\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,\qquad A\gt 0,\;B\gt 0.$$

Being inscribed means that every vertex of the rectangle lies on the ellipse, so $$\frac{2^{2}}{A^{2}}+\frac{1^{2}}{B^{2}}=1 \;\;\Longrightarrow\;\; \frac{4}{A^{2}}+\frac{1}{B^{2}}=1.\quad -(1)$$

We are further told that the outer ellipse passes through the point $$(4,0)$$. Substituting: $$\frac{4^{2}}{A^{2}}+0=1 \;\;\Longrightarrow\;\; \frac{16}{A^{2}}=1 \;\;\Longrightarrow\;\; A^{2}=16.\quad -(2)$$

Insert $$A^{2}=16$$ in equation $$-(1)$$: $$\frac{4}{16}+\frac{1}{B^{2}}=1 \;\;\Longrightarrow\;\; \frac{1}{4}+\frac{1}{B^{2}}=1 \;\;\Longrightarrow\;\; \frac{1}{B^{2}}=1-\frac{1}{4}=\frac{3}{4} \;\;\Longrightarrow\;\; B^{2}=\frac{4}{3}.$$

Thus the required ellipse is $$\frac{x^{2}}{16}+\frac{y^{2}}{4/3}=1.$$ Multiply by $$16$$ to remove denominators: $$x^{2}+16\left(\frac{y^{2}}{4/3}\right)=16 \;\;\Longrightarrow\;\; x^{2}+16\left(\frac{3}{4}y^{2}\right)=16 \;\;\Longrightarrow\;\; x^{2}+12y^{2}=16.$$

Hence the equation of the ellipse is $$x^{2}+12y^{2}=16$$.

Option B which is: $$x^{2}+12y^{2}=16$$