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અતિવલયના શિરોબિંદુઓ $(0, 0)$ અને $(10, 0)$ આગળ હોય અને તેની એક નાભિ $(18, 0)$ આગળ છે. અતિવલયનું સમીકરણ.....
$\frac{{{x^2}}}{{25}}\,\, - \,\,\frac{{{y^2}}}{{144}}\,\, = \,\,1$
$\frac{{{{\left( {x\,\, - \,\,5} \right)}^2}}}{{25}}\,\, - \,\,\frac{{{y^2}}}{{144}}\,\, = \,\,1$
$\frac{{{x^2}}}{{25}}\,\, - \,\,\frac{{{{\left( {y\,\, - \,\,5} \right)}^2}}}{{144}}\,\, = \,\,1$
$\frac{{{{\left( {x\,\, - \,\,5} \right)}^2}}}{{25}}\,\, - \,\,\frac{{{{\left( {y\,\, - \,\,5} \right)}^2}}}{{144}}\,\, = \,\,1$
Solution
$\,2a\,\, = \,\,10$ $a\,\, = \,\,5$
$ae\,\, – \,\,a\,\, = \,\,8\,\,\,$
$\,\,e\, = \,\,1\,\, + \;\,\frac{8}{5}\,\, = \,\,\frac{{13}}{5}\,$
$b\,\, = \,\,5\,\,\sqrt {\frac{{{{13}^2}}}{{{5^2}}}\,\, – \,\,1} \,\, = \,\,5\,\, \times \,\,\frac{{12}}{5}\,\, = \,\,12\,$
અને અતિવલય નું કેન્દ્ર $ \equiv \,\,\left( {5,\,\,0} \right)$
$\therefore \,\,\frac{{{{\left( {x\,\, – \,\,5} \right)}^2}}}{{{5^2}}}\,\, – \,\,\frac{{{{\left( {y\,\, – \,\,0} \right)}^2}}}{{{{12}^2}}}\,\, = \,\,1$
