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

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(D) The given equation of the ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$.The coordinates of the foci are $(\pm ae, 0) = (\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0) = (\pm 1, 0)$.The endpoints of the latus rectum are $(ae, \pm \frac{b^2}{a}) = (1, \pm \frac{1}{\sqrt{2}})$ and $(-1, \pm \frac{1}{\sqrt{2}})$.The equation of the tangent at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.For the point $(1, 1/\sqrt{2})$,the tangent is $\frac{x(1)}{2} + \frac{y(1/\sqrt{2})}{1} = 1$,which simplifies to $x + \sqrt{2}y = 2$.By symmetry,the four tangents are $x + \sqrt{2}y = 2$,$x - \sqrt{2}y = 2$,$-x + \sqrt{2}y = 2$,and $-x - \sqrt{2}y = 2$.These tangents form a rhombus. The vertices are found by solving the intersection of these lines: $(2, 0), (0, \sqrt{2}), (-2, 0), (0, -\sqrt{2})$.The area of the rhombus is $\frac{1}{2} 	imes 	ext{diagonal}_1 	imes 	ext{diagonal}_2 = \frac{1}{2} 	imes 4 	imes 2\sqrt{2} = 4\sqrt{2}$.

Find the area of the quadrilateral formed by the tangents at the endpoints of the latus rectum of the ellipse $x^2 + 2y^2 = 2$.

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