Find the equation of the ellipse (a b) whos | vedclass

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(D) Given that the distance between the foci is $2ae = 8$,so $ae = 4$.The distance between the directrices is $\frac{2a}{e} = 18$,so $\frac{a}{e} = 9$

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$5x^2 + 9y^2 = 180$

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(D) Given that the distance between the foci is $2ae = 8$,so $ae = 4$.The distance between the directrices is $\frac{2a}{e} = 18$,so $\frac{a}{e} = 9$.Multiplying these two equations: $(ae) 	imes (\frac{a}{e}) = 4 	imes 9$ $\Rightarrow a^2 = 36$ $\Rightarrow a = 6$.Dividing the equations: $\frac{ae}{a/e} = \frac{4}{9}$ $\Rightarrow e^2 = \frac{4}{9}$ $\Rightarrow e = \frac{2}{3}$.For an ellipse,$b^2 = a^2(1 - e^2) = 36(1 - \frac{4}{9}) = 36(\frac{5}{9}) = 20$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.Multiplying by $180$,we get $5x^2 + 9y^2 = 180$.

Find the equation of the ellipse $(a > b)$ whose distance between the foci is $8$ and the distance between the directrices is $18$.

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