Let the foci and length of the latus rectum of an ellipse [(x^2)/(a^2)+(y^2)/(b^2)=1, ab …

Question

Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$$ and $$\sqrt{50}$$

Accepted Answer

$\begin{gathered} \equiv (\pm 5,0);\frac{2{b}^{2}}{a}=\sqrt{50} \ a=5\, {b}^{2}=\frac{5\sqrt{2}a}{2} \ {b}^{2}={a}^{2}(1-{e}^{2})=\frac{5\sqrt{2}a}{2} \end{gathered}$

$\begin{gathered} \Rightarrow a(1-{e}^{2})=\frac{5\sqrt{2}}{2} \ \Rightarrow \frac{5}{e}(1-{e}^{2})=\frac{5\sqrt{2}}{2} \ \Rightarrow \sqrt{2}-\sqrt{2}{e}^{2}=e \ \Rightarrow \sqrt{2}{e}^{2}+e-\sqrt{2}=0 \ \Rightarrow \sqrt{2}{e}^{2}+2e-e-\sqrt{2}=0 \ \Rightarrow \sqrt{2}e(e+\sqrt{2})-1(1+\sqrt{2})=0 \ \Rightarrow (e+\sqrt{2})(\sqrt{2}e-1)=0 \ 	herefore e
e -\sqrt{2};e=\frac{1}{\sqrt{2}} \end{gathered}$

$\begin{gathered} \frac{{x}^{2}}{{b}^{2}}-\frac{{y}^{2}}{{a}^{2}{b}^{2}}=1\, a=5\sqrt{2} \ b=5 \ {a}^{2}{b}^{2}={b}^{2}({e}_{2}^{1}-1)\Rightarrow {e}_{2}^{1}=51 \end{gathered}$

Solutions · 1