If the eccentricity of the ellipse frac{x^2}{ | vedclass

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(B) Given the ellipse equation $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$. Since $a^2 + 2 > a^2 + 1$,the major axis is along the $y$-axis. Here,$

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$\frac{10}{\sqrt{6}}$

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(B) Given the ellipse equation $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$. Since $a^2 + 2 > a^2 + 1$,the major axis is along the $y$-axis. Here,$b^2 = a^2 + 1$ and $a_e^2 = a^2 + 2$ (where $a_e$ is the semi-major axis). The eccentricity $e = \frac{1}{\sqrt{6}}$,so $e^2 = \frac{1}{6}$. Using the relation $b^2 = a_e^2(1 - e^2)$: $a^2 + 1 = (a^2 + 2)(1 - \frac{1}{6})$ $a^2 + 1 = (a^2 + 2)(\frac{5}{6})$ $6a^2 + 6 = 5a^2 + 10$ $a^2 = 4$. Thus,$b^2 = 4 + 1 = 5$ and $a_e^2 = 4 + 2 = 6$. The length of the latus rectum is $\frac{2b^2}{a_e} = \frac{2(5)}{\sqrt{6}} = \frac{10}{\sqrt{6}}$.

If the eccentricity of the ellipse $\frac{x^2}{a^2 + 1} + \frac{y^2}{a^2 + 2} = 1$ is $\frac{1}{\sqrt{6}}$,find the length of the latus rectum of the ellipse.

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