The area of the quadrilateral formed by the t | vedclass

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

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$27$

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(D) For the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$,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 end points of the latus rectum are $(\pm ae, \pm \frac{b^2}{a}) = (\pm 2, \pm \frac{5}{3})$.The tangent at the point $(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$,or $\frac{x}{9/2} + \frac{y}{3} = 1$.This tangent forms a triangle with the coordinate axes in the first quadrant with intercepts $x_0 = 9/2$ and $y_0 = 3$.The area of this triangle is $\frac{1}{2} 	imes \frac{9}{2} 	imes 3 = \frac{27}{4}$.Since there are four such symmetric triangles formed by the tangents at the four ends of the latus recta,the total area of the quadrilateral is $4 	imes \frac{27}{4} = 27$ $sq. 	ext{ units}$.

The area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is .............. $sq. \text{ units}$.

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