Let S and S^{prime} be the foci of an ellipse | vedclass

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(C) The equation of the ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the

Vedclass · Accepted Answer

$80$

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(C) The equation of the ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.Given $\angle S^{\prime} SB = \frac{\pi}{6}$. In $	riangle OBS$,$\angle OSB = \frac{\pi}{6}$.Thus,$	an(\frac{\pi}{6}) = \frac{OB}{OS} = \frac{b}{ae} = \frac{1}{\sqrt{3}}$.This implies $3b^2 = a^2e^2$.Since $b^2 = a^2(1 - e^2)$,we have $a^2e^2 = a^2 - b^2$.Substituting this,$3b^2 = a^2 - b^2$,which gives $a^2 = 4b^2$.The point $(2\sqrt{3}, 1)$ lies on the ellipse,so $\frac{(2\sqrt{3})^2}{a^2} + \frac{1^2}{b^2} = 1$.Substituting $a^2 = 4b^2$,we get $\frac{12}{4b^2} + \frac{1}{b^2} = 1$,which simplifies to $\frac{3}{b^2} + \frac{1}{b^2} = 1$.Thus,$\frac{4}{b^2} = 1$,so $b^2 = 4$ and $a^2 = 16$.The length of the major axis is $2a$ and the length of the minor axis is $2b$.The sum of the squares of the lengths is $(2a)^2 + (2b)^2 = 4a^2 + 4b^2 = 4(16) + 4(4) = 64 + 16 = 80$.

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