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

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(B) Given the ellipse equation $\frac{x^2}{25}+\frac{y^2}{16}=1$,we have $a^2=25$ and $b^2=16$,so $a=5$ and $b=4$.Eccentricity $e = \sqrt{1 - \frac{b^

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$\frac{250}{3}$

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(B) Given the ellipse equation $\frac{x^2}{25}+\frac{y^2}{16}=1$,we have $a^2=25$ and $b^2=16$,so $a=5$ and $b=4$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.The ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 3, \pm \frac{16}{5})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.For the point $(3, \frac{16}{5})$,the tangent is $\frac{3x}{25} + \frac{(16/5)y}{16} = 1$,which simplifies to $\frac{3x}{25} + \frac{y}{5} = 1$.This line intersects the $x$-axis at $P(\frac{25}{3}, 0)$ and the $y$-axis at $Q(0, 5)$.The area of the triangle $OPQ$ in the first quadrant is $\frac{1}{2} 	imes \frac{25}{3} 	imes 5 = \frac{125}{6}$.By symmetry,the total area of the quadrilateral formed by the four tangents is $4 	imes \frac{125}{6} = \frac{250}{3}$ sq units.

Tangents are drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at all the four ends of its latus rectum. The area (in sq units) of the quadrilateral formed by these tangents is

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