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Q.
If the eccentricity of the ellipse $\frac{x^{2}}{a^{2}+1}+\frac{y^{2}}{a^{2}+2}=1$ is $1 / \sqrt{6}$, then the latus rectum of the ellipse is
Conic Sections
Solution:
Here, $a^{2}+2>a^{2}+1$
or $a^{2}+1=\left(a^{2}+2\right)\left(1-e^{2}\right)$
or $ a^{2}+1=\left(a^{2}+2\right) \frac{5}{6}$
or $ 6 a^{2}+6=5 a^{2}+10$
or $a^{2}=10-6=4$
or $a=\pm 2$
Latus rectum $=\frac{2\left(a^{2}+1\right)}{\sqrt{a^{2}+2}}=\frac{2 \times 5}{\sqrt{6}}=\frac{10}{\sqrt{6}}$