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બિંદુ $\left( {a\,\,\sec \,\theta ,\,\,b\,\,\tan \,\,\theta } \right)$ આગળ અતિવલય $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,$ ના અભિલંબનું સમીકરણ મેળવો.
$\frac{{ax}}{{\sec \,\,\theta }}\,\, - \,\,\frac{{by}}{{\tan \,\,\theta }}\,\, = \,\,{a^2}\,\, - \,\,{b^2}$
$\frac{{ax}}{{\sec \,\,\theta }}\,\, + \,\,\frac{{by}}{{\tan \,\,\theta }}\,\, = \,\,{a^2}\,\, + \,\,{b^2}$
$\frac{{ax}}{{\sec \,\,\theta }}\,\, + \,\frac{{by}}{{\tan \,\,\theta }}\,\, = \,\,{a^2}\,\, - \,\,{b^2}$
$\frac{{ax}}{{\sec \,\,\theta }}\,\, - \,\,\frac{{by}}{{\tan \,\,\theta }}\,\, = \,\,a\,\, - \,b$
Solution
.$\,\,\,\left( {a\,\sec \,\,\theta ,\,\,b\,\tan \,\theta } \right)\,\,$ આગળ અતિવલય $\frac{{{x^2}}}{{{a^2}}}\,\, – \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1$ ના
અભિલ્ંબનું સમીકરણ $\,\,\frac{{{a^2}x}}{{a\,\,\sec \,\,\theta }}\,\, + \;\,\frac{{{b^2}y}}{{b\,\,\tan \,\,\theta }}\,\, = \,\,{a^2}\,\, + \;\,{b^2}$
