The coordinates of a point,in the parametric | vedclass

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(B) The center of the ellipse is the midpoint of the foci $(-1, 0)$ and $(7, 0)$,which is $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.The distance betw

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$(3 + 8 \cos 	heta, 4 \sqrt{3} \sin 	heta)$

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(B) The center of the ellipse is the midpoint of the foci $(-1, 0)$ and $(7, 0)$,which is $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.The distance between the foci is $2ae = 7 - (-1) = 8$,so $ae = 4$.Given $e = \frac{1}{2}$,we have $a(\frac{1}{2}) = 4$,which implies $a = 8$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 8^2(1 - (\frac{1}{2})^2) = 64(1 - \frac{1}{4}) = 64(\frac{3}{4}) = 48$.Thus,$b = \sqrt{48} = 4 \sqrt{3}$.The parametric coordinates of a point on an ellipse with center $(h, k)$ are $(h + a \cos 	heta, k + b \sin 	heta)$.Substituting the values,we get $(3 + 8 \cos 	heta, 0 + 4 \sqrt{3} \sin 	heta) = (3 + 8 \cos 	heta, 4 \sqrt{3} \sin 	heta)$.

The coordinates of a point,in the parametric form,on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity $e = \frac{1}{2}$,are

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