The ellipse 4x^2 + 9y^2 = 1 intersects the po | vedclass

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(A) The equation of the ellipse is $4x^2 + 9y^2 = 1$, which can be written as $\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.Here, $a^2 = 1/4$ and $b

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$\frac{\sqrt{5}}{18}$

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(A) The equation of the ellipse is $4x^2 + 9y^2 = 1$, which can be written as $\frac{x^2}{(1/2)^2} + \frac{y^2}{(1/3)^2} = 1$.Here, $a^2 = 1/4$ and $b^2 = 1/9$. Since $a^2 > b^2$, the major axis is along the $X$-axis.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/9}{1/4}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.The foci $S$ and $S'$ are at $(\pm ae, 0)$, so $S = (\frac{1}{2} \cdot \frac{\sqrt{5}}{3}, 0) = (\frac{\sqrt{5}}{6}, 0)$ and $S' = (-\frac{\sqrt{5}}{6}, 0)$.The ellipse intersects the positive $Y$-axis at $A(0, b) = (0, 1/3)$.The base of $\Delta SAS'$ is the distance $SS' = \frac{\sqrt{5}}{6} - (-\frac{\sqrt{5}}{6}) = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.The height of the triangle is the $y$-coordinate of $A$, which is $1/3$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{\sqrt{5}}{3} 	imes \frac{1}{3} = \frac{\sqrt{5}}{18}$ sq. units.

The ellipse $4x^2 + 9y^2 = 1$ intersects the positive $Y$-axis at point $A$. If $S$ and $S'$ are its foci, then the area (in sq. units) of $\Delta SAS'$ is

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