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10-2. Parabola, Ellipse, Hyperbola
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અનુપ્રસ્થ અક્ષોની લંબાઈ $2\ sin\ \theta$ ધરાવતો અતિવલય, એ ઉપવલય $3x^2 + 4y^2 = 12$ સાથે સમનાભિ હોય, તો તેનું સમીકરણ.....

A

$x^2cosec^2\theta - y^2sec^2\theta = 1$

B

$x^2sec^2\theta - y^2cosec^2\theta = 1$

C

$x^2sin^2\theta - y^2cos^2\theta = 1$

D

$x^2cos^2\theta - y^2sin^2\theta = 1$

Solution

આપેલો ઉપવલય ${\rm{3}}{{\rm{x}}^{\rm{2}}}\,\, + \;\,4{y^2}\,\, = \,\,12$

$\therefore \,\,{a^2}\,\, = \,\,4\,\,\,\,{b^2}\,\, = \,\,3$

$\therefore \,\,e\,\, = \,\,\sqrt {1\,\, – \,\,\frac{3}{4}} \,\,\, \Rightarrow \,\,e\,\, = \,\,\frac{1}{2}\,$

$\therefore \,$ તેનું નાભિકેન્દ્ર $\left( { \pm \,1,\,\,0} \right)$  થશે.

જ્યારે અતિવલય એ આપેલા ઉપવલય ને સમાનભિય હોય , તેથી $ \pm \,\,ae\,\, = \,\, \pm 1,\,\,$  પરંતુ $a\,\, = \,\,\sin \,\theta $  આપેલ છે.

$\therefore \,\,e\,\, = \,\,\frac{1}{{\sin \,\,\theta }},$ હવે ${b^2}\,\, = \,\,{a^2}\,\,\left( {{e^2}\,\, – \,\,1} \right)$

${b^2}\, = \,\,{\sin ^2}\,\,\theta \,\,\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\,\, \Rightarrow \,\,{b^2}\,\, = \,\,{\cos ^2}\theta $

આથી માંગેલું સમીકરણ $ :\,\,\frac{{{x^2}}}{{{{\sin }^2}\theta }}\,\, – \,\,\frac{{{y^2}}}{{{{\cos }^2}\,\,\theta }}\,\, = \,\,1$

$\, \Rightarrow \,\,{x^2}\,\cos e{c^2}\theta \,\, – \,\,{y^2}\,{\sec ^2}\theta \,\, = \,\,1$

Standard 11
Mathematics

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