The length of the latus-rectum of the ellipse | vedclass

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(D) The distance between the foci is $2ae = \sqrt{(2-2)^2 + (5 - (-3))^2} = \sqrt{0^2 + 8^2} = 8$.Given the eccentricity $e = \frac{4}{5}$,we have $2a

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$\frac{18}{5}$

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(D) The distance between the foci is $2ae = \sqrt{(2-2)^2 + (5 - (-3))^2} = \sqrt{0^2 + 8^2} = 8$.Given the eccentricity $e = \frac{4}{5}$,we have $2ae = 8$,so $ae = 4$.Substituting $e = \frac{4}{5}$,we get $a 	imes \frac{4}{5} = 4$,which implies $a = 5$.For an ellipse,the relation between the semi-major axis $a$,semi-minor axis $b$,and the distance from the center to the focus $ae$ is $a^2 = b^2 + (ae)^2$ (where $a > b$).Here,$a = 5$ and $ae = 4$,so $5^2 = b^2 + 4^2$.$25 = b^2 + 16$,which gives $b^2 = 9$,so $b = 3$.The length of the latus-rectum is given by $\frac{2b^2}{a}$.Substituting the values,$	ext{L.R.} = \frac{2 	imes 9}{5} = \frac{18}{5}$.

The length of the latus-rectum of the ellipse,whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$,is

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