The ellipse x^2 + 4y^2 = 4 is inscribed in a | vedclass

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(A) The given ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.This ellipse has semi-axes $a = 2$ and $b = 1$.T

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$x^2 + 12y^2 = 16$

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(A) The given ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.This ellipse has semi-axes $a = 2$ and $b = 1$.The rectangle inscribed around this ellipse has vertices at $(\pm 2, \pm 1)$.The outer ellipse is inscribed in this rectangle,meaning it passes through the points $(\pm 2, \pm 1)$.Let the equation of the outer ellipse be $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.Since it passes through $(4,0)$,we have $\frac{4^2}{A^2} + \frac{0^2}{B^2} = 1$,which gives $A^2 = 16$.Since it also passes through $(2,1)$,we have $\frac{2^2}{16} + \frac{1^2}{B^2} = 1$.$\frac{4}{16} + \frac{1}{B^2} = 1 \implies \frac{1}{4} + \frac{1}{B^2} = 1 \implies \frac{1}{B^2} = \frac{3}{4} \implies B^2 = \frac{4}{3}$.Substituting $A^2$ and $B^2$ into the equation: $\frac{x^2}{16} + \frac{y^2}{4/3} = 1$.$\frac{x^2}{16} + \frac{3y^2}{4} = 1$.Multiplying by $16$,we get $x^2 + 12y^2 = 16$.

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 outer ellipse is:

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