If tangents are drawn to the ellipse frac{x^2 | vedclass

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(A) The given equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$. Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we have $a^2=9$ and $b^2=

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$27$ sq. units

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(A) The given equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$. Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,we have $a^2=9$ and $b^2=5$. 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 coordinates of the foci are $(\pm ae, 0) = (\pm 3 \cdot \frac{2}{3}, 0) = (\pm 2, 0)$. The endpoints of the latus recta are $L(2, \frac{5}{3})$,$M(-2, \frac{5}{3})$,$M'(-2, -\frac{5}{3})$,and $L'(2, -\frac{5}{3})$. 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 $L(2, \frac{5}{3})$,the tangent is $\frac{2x}{9}+\frac{5y/3}{5}=1$ $\Rightarrow \frac{2x}{9}+\frac{y}{3}=1$ $\Rightarrow 2x+3y=9$. Similarly,the other tangents are $2x-3y=9$,$-2x+3y=9$,and $-2x-3y=9$. These lines form a rhombus with vertices at $A(0, 3)$,$B(-\frac{9}{2}, 0)$,$C(0, -3)$,and $D(\frac{9}{2}, 0)$. The area of the rhombus is $\frac{1}{2} 	imes 	ext{diagonal}_1 	imes 	ext{diagonal}_2 = \frac{1}{2} 	imes (2 \cdot \frac{9}{2}) 	imes (2 \cdot 3) = \frac{1}{2} 	imes 9 	imes 6 = 27$ sq. units.

If tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at the ends of the latus recta,then the area of the quadrilateral thus formed is

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