A) The equation of the hyperbola is \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\]
B) A focus of the hyperbola is \[(\sqrt{3},0)\]
C) The eccentricity of the hyperbola is \[\sqrt{\frac{5}{3}}\]
D) The equation of the hyperbola is \[{{x}^{2}}-3{{y}^{2}}=3\]
Correct Answer: D
Solution :
[d] Given ellipse is \[\frac{{{x}^{4}}}{4}+\frac{{{y}^{2}}}{1}=1.\] Eccentricity of the ellipse \[=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\] Focus of the ellipse \[=(-\sqrt{3},\,0)\] Eccentricity of the hyperbola \[=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] So, \[\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{2}{\sqrt{3}}\] \[\Rightarrow \,\,\frac{b}{a}=\frac{1}{\sqrt{3}}\] Since the hyperbola passes through the focus of the ellipse, \[\frac{3}{{{a}^{2}}}-0=1\] \[\Rightarrow \,\,\,\,{{a}^{2}}=3\] and \[{{b}^{2}}=1\] Thus, the equation of the hyperbola is \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{1}=1\] or \[{{x}^{2}}-3{{y}^{2}}=3.\] Focus of hyperbola is \[(\pm \,2,\,0)\].
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