question_answer

A hyperbola whose transverse axis is along the major axis of the conic,\[\frac{{{x}^{2}}}{3}+\frac{{{y}^{2}}}{4}=4\] and has vertices at the foci of this conic. If the eccentricity of the hyperbola is\[\frac{3}{2}\], then which of the following points does NOT lie on it?   JEE Main Online Paper (Held On 10 April 2016)

A) \[\left( \sqrt{5},2\sqrt{2} \right)\]                              

B) \[\left( 5,2\sqrt{3} \right)\]

C) \[\left( 0,\,\,2 \right)\]     

D) \[\left( \sqrt{10},\,\,2\sqrt{3} \right)\]

Correct Answer: C

Solution :

             Ellipse\[\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{16}=1\] foci \[(0,\pm be)\] \[{{e}_{e}}=\sqrt{1-\frac{12}{16}}=\frac{1}{2}\]= for hyperbola \[{{h}_{h}}=2\] \[{{e}_{H}}=\frac{3}{2}\] equation \[\Rightarrow \,\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] \[{{e}_{H}}\frac{3}{2}=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}\]            \[\Rightarrow \]               \[\frac{9}{4}-1=\frac{{{a}^{2}}}{{{b}^{2}}}\] \[\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{5}{4}\]                 \[\Rightarrow \]               \[{{a}^{2}}=5\]              \[\frac{{{x}^{2}}}{5}-\frac{{{y}^{2}}}{4}=-1\]