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જો $\alpha$ અને $\beta$ એ ઉપવલય $\frac{{{x^2}}}{{{a^2}}}\,\, + \;\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1$ની નાભિજીવાના અંત્યબિંદુઓના ઉત્કેન્દ્રીકરણ હોય, તો $tan\ \alpha /2. tan\ \beta/2 = ....$
$\frac{{e\,\, - \,\,1}}{{e\,\, + \;\,1}}$
$\frac{{1\,\, - \,\,e}}{{1\,\, + \;\,e}}$
$\frac{{e\,\, + \,\,1}}{{e\,\, - \;\,1}}$
$\frac{{e\,\, - \,\,1}}{{1\,\, + \;\,e}}$
Solution
$'\alpha '$ અને $'\beta '\,$ બિંદુઓને જોડતી રેખાનું સમીકરણ
$\frac{x}{a}\,\,\cos \,\,\frac{{\alpha \,\, + \;\,\beta }}{2}\,\, + \,\,\frac{y}{b}\sin \,\,\frac{{\alpha \,\, + \;\,\beta }}{2}\,\, = \,\,\cos \,\,\frac{{\alpha \,\, – \,\,\beta }}{2}$
જો તે નાભિજીવા હોય, તો તે નાભિકેન્દ્ર $\left( {ae,\,\,0} \right)\,$ માંથી પસાર થાય,તેથી
$e\,\,\cos \,\,\frac{{\alpha \,\, + \;\,\beta }}{2}\,\, = \,\cos \,\frac{{\alpha \,\, – \,\,\beta }}{2}\,\, \Rightarrow \,\,\,\frac{{\cos \,\,\frac{{\alpha \,\, – \,\,\beta }}{2}}}{{\cos \,\,\frac{{\alpha \,\, + \;\,\beta }}{2}}}\,\, = \,\,\frac{e}{1}\,\,$
$ \Rightarrow \,\,\frac{{\cos \,\,\frac{{\alpha \,\, – \,\,\beta }}{2}\,\, – \,\,\cos \,\,\,\frac{{\alpha \, + \,\,\beta }}{2}}}{{\cos \,\,\frac{{\alpha \,\, – \,\,\beta }}{2}\,\, + \;\,\cos \,\,\frac{{\alpha \,\, + \;\,\beta }}{2}}}\,\, = \,\,\frac{{e\,\, – \,\,1}}{{e\,\, + \;\,1}}\,\,$
$ \Rightarrow \,\,\frac{{2\,\sin \,\,\alpha /2\,\,\sin \,\,\beta /2}}{{2\,\,\cos \,\,\alpha /2\,\,\cos \,\,\beta /2}}\,\, = \,\,\frac{{e\,\, – \,\,1}}{{e\,\, + \;\,1}}\,\, \Rightarrow \,\,\tan \,\,\frac{\alpha }{2}\,\,\tan \,\,\frac{\beta }{2}\,\, = \,\,\frac{{e\,\, – \,\,1}}{{e\,\, + \;\,1}}$
