The parametric equations of the ellipse whose | vedclass

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(B) The foci are given as $F_1(-3, 0)$ and $F_2(9, 0)$.Since the $y$-coordinates are the same,the major axis is horizontal.The center $(h, k)$ is the

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$x=3+18 \cos 	heta, y=12 \sqrt{2} \sin 	heta$

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(B) The foci are given as $F_1(-3, 0)$ and $F_2(9, 0)$.Since the $y$-coordinates are the same,the major axis is horizontal.The center $(h, k)$ is the midpoint of the foci: $h = \frac{-3+9}{2} = 3$ and $k = 0$.The distance between the foci is $2ae = 9 - (-3) = 12$,so $ae = 6$.Given eccentricity $e = \frac{1}{3}$,we have $a(\frac{1}{3}) = 6$,which gives $a = 18$.Using the relation $b^2 = a^2(1-e^2)$,we get $b^2 = 18^2(1 - (\frac{1}{3})^2) = 324(1 - \frac{1}{9}) = 324(\frac{8}{9}) = 288$.Thus,$b = \sqrt{288} = 12\sqrt{2}$.The parametric equations for a shifted ellipse are $x = h + a \cos 	heta$ and $y = k + b \sin 	heta$.Substituting the values,we get $x = 3 + 18 \cos 	heta$ and $y = 12\sqrt{2} \sin 	heta$.

The parametric equations of the ellipse whose foci are $(-3, 0)$ and $(9, 0)$ and eccentricity is $\frac{1}{3}$,are

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