An ellipse passing through (4 sqrt{2}, 2 sqrt | vedclass

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(B) The foci are at $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since $b^2 = a^2(1 -

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$\frac{1}{2}$

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(B) The foci are at $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since $b^2 = a^2(1 - e^2) = a^2 - a^2e^2 = a^2 - 16$,the equation becomes $\frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} = 1$.Substituting the point $(4 \sqrt{2}, 2 \sqrt{6})$:$\frac{(4 \sqrt{2})^2}{a^2} + \frac{(2 \sqrt{6})^2}{a^2 - 16} = 1$$\frac{32}{a^2} + \frac{24}{a^2 - 16} = 1$$32(a^2 - 16) + 24a^2 = a^2(a^2 - 16)$$56a^2 - 512 = a^4 - 16a^2$$a^4 - 72a^2 + 512 = 0$$(a^2 - 64)(a^2 - 8) = 0$.Since $a > ae = 4$,$a^2$ must be $64$,so $a = 8$.Then $e = \frac{ae}{a} = \frac{4}{8} = \frac{1}{2}$.

Molecule type	$\frac{C_P}{C_V}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

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