The eccentricity of an ellipse passing throug | vedclass

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(3 \sqrt{2}, \sqrt{10})$,we have $\fr

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$\frac{2}{3}$

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(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(3 \sqrt{2}, \sqrt{10})$,we have $\frac{(3 \sqrt{2})^2}{a^2} + \frac{(\sqrt{10})^2}{b^2} = 1$,which simplifies to $\frac{18}{a^2} + \frac{10}{b^2} = 1$ $(i)$.The foci are at $(\pm 4, 0)$,so $ae = 4$,which means $a^2 e^2 = 16$.Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2 e^2$,we get $b^2 = a^2 - 16$.Substituting $b^2$ into equation $(i)$: $\frac{18}{a^2} + \frac{10}{a^2 - 16} = 1$.Multiplying by $a^2(a^2 - 16)$,we get $18(a^2 - 16) + 10a^2 = a^2(a^2 - 16)$.$18a^2 - 288 + 10a^2 = a^4 - 16a^2$.$a^4 - 44a^2 + 288 = 0$.Factoring the quadratic in $a^2$: $(a^2 - 36)(a^2 - 8) = 0$.Since $b^2 = a^2 - 16 > 0$,we must have $a^2 > 16$,so $a^2 = 36$.Finally,$e^2 = \frac{16}{a^2} = \frac{16}{36} = \frac{4}{9}$.Therefore,$e = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The eccentricity of an ellipse passing through $(3 \sqrt{2}, \sqrt{10})$ with foci at $(-4,0)$ and $(4,0)$ is

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