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

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(D) Given the ellipse equation is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$,so $a = 2$

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

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(D) Given the ellipse equation is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$.Here,$a^2 = 4$ and $b^2 = 1$,so $a = 2$ and $b = 1$.The rectangle inscribed in this ellipse has vertices at $(\pm a, \pm b)$,which are $(\pm 2, \pm 1)$.The sides of this rectangle are parallel to the coordinate axes.We need to find the equation of an ellipse that circumscribes this rectangle. Let the equation be $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.Since the rectangle is inscribed in the new ellipse,the vertices $(\pm 2, \pm 1)$ must satisfy the equation.Thus,$\frac{2^2}{A^2} + \frac{1^2}{B^2} = 1$,which simplifies to $\frac{4}{A^2} + \frac{1}{B^2} = 1$.For an ellipse to circumscribe a rectangle with sides parallel to the axes,the ratio of the semi-axes must be proportional to the dimensions of the rectangle. Thus,$\frac{A}{B} = \frac{2}{1} = 2$,so $A = 2B$.Substituting $A = 2B$ into the equation: $\frac{4}{(2B)^2} + \frac{1}{B^2} = 1$.$\frac{4}{4B^2} + \frac{1}{B^2} = 1 \implies \frac{1}{B^2} + \frac{1}{B^2} = 1 \implies \frac{2}{B^2} = 1 \implies B^2 = 2$.Then $A^2 = (2B)^2 = 4B^2 = 4(2) = 8$.The equation is $\frac{x^2}{8} + \frac{y^2}{2} = 1$,which is $x^2 + 4y^2 = 8$. However,checking the options,let's re-evaluate the condition. If the rectangle is inscribed in the original ellipse,its sides are $x = \pm 2$ and $y = \pm 1$. The ellipse circumscribing this rectangle must pass through these points. If we assume the ellipse is of the form $x^2 + ky^2 = c$,then $4 + k = c$ and $0 + 4k = c$ is not correct. Let's test option $C$: $x^2 + 16y^2 = 16 \implies \frac{x^2}{16} + \frac{y^2}{1} = 1$. At $(\pm 2, \pm 1)$,$\frac{4}{16} + \frac{1}{1} = \frac{1}{4} + 1 = 1.25 
eq 1$. Let's re-read: the rectangle is formed by the tangents at the vertices of the ellipse? No,it says inscribed. The vertices of the rectangle are $(\pm 2, \pm 1)$. The ellipse circumscribing it is $x^2 + 4y^2 = 8$. Since this is not an option,let's check $x^2 + 12y^2 = 16 \implies \frac{x^2}{16} + \frac{y^2}{4/3} = 1$. At $(\pm 2, \pm 1)$,$\frac{4}{16} + \frac{1}{4/3} = \frac{1}{4} + \frac{3}{4} = 1$. This satisfies the condition. Thus,option $D$ is correct.

The ellipse $x^2 + 4y^2 = 4$ is inscribed in a rectangle whose sides are parallel to the coordinate axes. Find the equation of the ellipse that circumscribes this rectangle.

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