Question

# Find value of $\left(5-{e}^{2}\right)$ for ellipse $\frac{{x}^{2}}{8}+\frac{{y}^{2}}{4}=1$

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Solution

## To find : $\left(5-{e}^{2}\right)$Given: the equation of an ellipse, $\frac{{x}^{2}}{8}+\frac{{y}^{2}}{4}=1$We know, for an ellipse$\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{}\mathbit{a}\mathbf{>}\mathbit{b}$an eccentricity is given by $\mathbit{e}\mathbf{=}\sqrt{\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}$So, for the given equation of an ellipse,$\beta e=\sqrt{\frac{{a}^{2}-{b}^{2}}{{a}^{2}}}\phantom{\rule{0ex}{0ex}}\beta e=\sqrt{\frac{8-4}{8}}\phantom{\rule{0ex}{0ex}}\beta e=\sqrt{\frac{1}{2}}$So, the value of $\left(5-{e}^{2}\right)$ would be,$\beta 5-{e}^{2}=5-{\left(\sqrt{\frac{1}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\beta 5-{e}^{2}=5-\frac{1}{2}\phantom{\rule{0ex}{0ex}}\beta \mathbf{5}\mathbf{-}{\mathbit{e}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}$Hence, $\mathbf{5}\mathbf{-}{\mathbit{e}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}$

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