A) \[\sqrt{\frac{7}{4}}\]
B) \[\sqrt{\frac{7}{3}}\]
C) \[\sqrt{\frac{5}{4}}\]
D) \[\sqrt{\frac{5}{3}}\]
Correct Answer: B
Solution :
Let equation of hyperbola and ellipse be \[\tan \,\phi =\frac{V}{H}\] ?(i) and \[\theta \] ?(ii) \[H'=H\,\cos \,\theta ,\] \[\phi '\] (since, same pair of foci) \[\tan \,\phi =\frac{V}{{{H}^{.}}}=\frac{V}{H\,\cos \,\theta }\] (given) Both intersect at (2, 2), \[\frac{\tan \,\phi '}{\tan \,\phi }=\frac{V/H\,\cos \,\theta }{\frac{V}{H}}=\frac{1}{\cos \,\theta }\] [from Eq. (ii)] \[{{\phi }_{1}}=\] \[=BA\] \[{{\phi }_{2}}=\] \[||\] Also, \[=0\] \[\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}\] \[(\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}})\] ?(iii) Now, \[T=\frac{2\pi }{\omega }\] [From Eq. (i)] \[\Rightarrow \] \[f'(x)=2+\cos \,x>0\] \[x\] \[\therefore \] \[f(x)\] \[\therefore \] \[f\] \[|M|=\alpha \] or \[{{M}^{-1}}adjM)=kI\] \[\Rightarrow \] \[|{{M}^{-1}}|\,|adj\,(adjM)|\,=\,|kI|\] \[\Rightarrow \] [From Eq. (iii)]
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