Tangents are drawn to the ellipse frac{x^{2}} | vedclass

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(A) Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, we have $a^{2}=9$ and $b^{2}=5$, so $a=3$ and $b=\sqrt{5}$.The eccentricity $e$ is given by

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$27 	ext{ sq units}$

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(A) Given the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$, we have $a^{2}=9$ and $b^{2}=5$, so $a=3$ and $b=\sqrt{5}$.The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.The foci are at $(\pm ae, 0) = (\pm 3 	imes \frac{2}{3}, 0) = (\pm 2, 0)$.The ends of the latus recta are at $(\pm 2, \pm \frac{b^{2}}{a}) = (\pm 2, \pm \frac{5}{3})$.Consider the point $P(2, \frac{5}{3})$ in the first quadrant. The equation of the tangent at $P$ is $\frac{x(2)}{9} + \frac{y(5/3)}{5} = 1$, which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$, or $2x + 3y = 9$.This line intersects the $x$-axis at $A(\frac{9}{2}, 0)$ and the $y$-axis at $B(0, 3)$.The quadrilateral formed by the four tangents is symmetric about both axes. The area of the quadrilateral is $4 	imes 	ext{Area}(\Delta OAB) = 4 	imes (\frac{1}{2} 	imes OA 	imes OB) = 4 	imes \frac{1}{2} 	imes \frac{9}{2} 	imes 3 = 27 	ext{ sq units}$.

Tangents are drawn to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ at the ends of both latus recta. The area of the quadrilateral so formed is

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