Tangents are drawn to the ellipse frac{x^2}{9 | vedclass

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(A) For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a})$.Given $\frac{x^2}{9}+\frac{y^2}

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$27$

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(A) For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the ends of the latus rectum are $(\pm ae, \pm \frac{b^2}{a})$.Given $\frac{x^2}{9}+\frac{y^2}{5}=1$,we have $a^2=9, b^2=5$.$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{5}{9} = \frac{4}{9} \Rightarrow e = \frac{2}{3}$.Ends of latus rectum are $(\pm 2, \pm \frac{5}{3})$.The equation of the tangent at $P(2, \frac{5}{3})$ is $\frac{2x}{9} + \frac{5y/3}{5} = 1$,which simplifies to $\frac{2x}{9} + \frac{y}{3} = 1$.This line intersects the axes at $A(\frac{9}{2}, 0)$ and $C(0, 3)$.The area of the triangle formed in the first quadrant 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 rectum,the total area is $4 	imes \frac{27}{4} = 27$ sq. units.

Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at all the ends of its latus rectum. The area of the quadrilateral so formed (in sq. units) is

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