The parametric form of a point on the ellipse | vedclass

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(B) The center of the ellipse is the midpoint of the foci: $(\frac{-1+7}{2}, \frac{0+0}{2}) = (3, 0)$.The distance between the foci is $2ae = \sqrt{(7

Vedclass · Accepted Answer

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

Vedclass · Answer

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

The parametric form of a point on the ellipse whose foci are $(-1, 0)$ and $(7, 0)$ and eccentricity is $1/2$ is:

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