Let x2a2+y2b2=1(a>b) be a given ellipse.Suppose an ellipse congruent to the given ellipse is drawn with the same centre and with the same axes such that it touches the given ellipse at the extremities of the major axis and having its foci on the extremities of minor axis of the given ellipse.Find the semi axes and eccentricity of the new ellipse.

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man111 singh ·

\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 $\bf{(\pm c,0) = (\pm a,0)}$

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