**·**2013-10-25 10:11:34

\hspace{-16}$Here original equation of ellipse $\bf{E_{1}::\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}.(a>b)$\\\\\\ Let $\bf{e}$ be the eccentricity of $\bf{E_{1}}$ and $\bf{C}$ be a center and $\bf{A(a,0)}$ and $\bf{A^{'}(-a,0)}$\\\\\\ be the extermeties of major - axis. and $\bf{B(0,b)}$ and $\bf{B^{'}(0,-b)}$ are\\\\\\ extermeties of minor-axis.\\\\\\ Now let $\bf{E_{2}::\frac{x^2}{c^2}+\frac{y^2}{d^2} = 1.\left(c<d\right)}$\\\\\\ Let $\bf{e^{'}}$ be the eccentricity of $\bf{E_{2}}$.\\\\\\ Now for Ellipse $\bf{E_{1}::}$\\\\\\ $\bf{b^2=a^2\left(1-e^2\right)}$\\\\\\ Now for Ellipse $\bf{E_{2}::}$\\\\\\ Ellipse passes through $\bf{\left(\pm a,0\right)}$ . So $\bf{c=a}$\\\\\\ Now ellipse $\bf{E_{2}}$ has a focus at $\bf{(0,\pm de^{'})}$\\\\\\ Which coincide with minor -axis of $\bf{E_{1}}$ which is $\bf{\left(0,\pm b\right)}$\\\\\\

\hspace{-16}$So $\bf{de^{'}=b\Rightarrow d = \frac{b}{e^{'}}}$\\\\\\ Now $\bf{c^2=d^2(1-e^{'}^2)\Rightarrow a^2=\frac{b^2}{e^{'}^2}\left(1-e^{'}^2\right)}$\\\\\\ So $\bf{e^{'}^2 = \frac{b^2}{a^2+b^2}\Rightarrow e^{'}=\frac{b}{\sqrt{a^2+b^2}}}$\\\\\\ So extermeties of major- axis and minir - axis of $\bf{E_{2}}$ are respectively\\\\\\ $\bf{(0,\pm d)=\left(0,\pm \sqrt{a^2+b^2}\right)}$ and $\bf{(\pm c,0) = (\pm a,0)}$

- Anik Chatterjee thanks

·0·Reply·2013-10-27 01:40:09