Question

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

Answer 1

To find : $(5-e^2)$

Given: the equation of an ellipse, $\frac{x^2}{8}+\frac{y^2}{4}=1$

We know, for an ellipse$\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{}\mathit{a}\mathbf{>}\mathit{b}$an eccentricity is given by $\mathit{e}\mathbf{=}\sqrt{\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}$

So, for the given equation of an ellipse,

$$\begin{gathered} \Rightarrow e=\sqrt{\frac{a^2-b^2}{a^2}} \\ \Rightarrow e=\sqrt{\frac{8-4}{8}} \\ \Rightarrow e=\sqrt{\frac{1}{2}} \end{gathered}$$

So, the value of $(5-e^2)$ would be,

$$\begin{gathered} \Rightarrow 5-e^2=5-{\left(\sqrt{\frac{1}{2}}\right)}^2 \\ \Rightarrow 5-e^2=5-\frac{1}{2} \\ \Rightarrow \mathbf{5}\mathbf{-}{\mathit{e}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}} \end{gathered}$$

Hence, $\mathbf{5}\mathbf{-}{\mathit{e}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}$