The coordinates of any point,in the parametri | vedclass

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(B) The foci are $(-2,0)$ and $(8,0)$.The distance between the foci is $2ae = 8 - (-2) = 10$.$\Rightarrow ae = 5$.Given $e = \frac{1}{\sqrt{2}}$,we ha

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$(3+5 \sqrt{2} \cos 	heta, 5 \sin 	heta)$

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(B) The foci are $(-2,0)$ and $(8,0)$.The distance between the foci is $2ae = 8 - (-2) = 10$.$\Rightarrow ae = 5$.Given $e = \frac{1}{\sqrt{2}}$,we have $a \left(\frac{1}{\sqrt{2}}ight) = 5 \Rightarrow a = 5\sqrt{2}$.Now,$b^2 = a^2(1 - e^2) = (5\sqrt{2})^2 \left(1 - \frac{1}{2}ight) = 50 \left(\frac{1}{2}ight) = 25$.Thus,$b = 5$.The centre of the ellipse is the mid-point of the line joining the two foci: $\left(\frac{-2+8}{2}, \frac{0+0}{2}ight) = (3,0)$.The equation of the ellipse is $\frac{(x-3)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$,which is $\frac{(x-3)^2}{50} + \frac{y^2}{25} = 1$.The parametric coordinates are $(x, y) = (h + a \cos 	heta, k + b \sin 	heta)$,where $(h, k)$ is the centre.Substituting the values,we get $(3 + 5\sqrt{2} \cos 	heta, 5 \sin 	heta)$.

The coordinates of any point,in the parametric form,on the ellipse whose foci are $(-2,0)$ and $(8,0)$ and eccentricity is $\frac{1}{\sqrt{2}}$,is

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