A) \[\sqrt{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) 2
D) 1
Correct Answer: A
Solution :
Equation of conics \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{4}=1,\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{9}=1\] Equation of eccentricity of an ellipse \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[4=16(1-{{e}^{2}})\] \[\Rightarrow \] \[{{e}^{2}}=1-1/4=3/4\] \[e=\pm \frac{\sqrt{3}}{4}\] Focii of an ellipse \[=(\pm \,ae,\,0)\] \[=\left( \pm 4.\frac{\sqrt{3}}{2},0 \right)=(\pm \,2\sqrt{3},0)\] Given, focii of both conics are coincides. \[\Rightarrow \] \[(\pm \,2\sqrt{3},0)=(\pm \,ae,\,0)\] [\[\because \] Here \[(\pm \,ae,0)\]is focii of second conic.] \[\Rightarrow \] \[\pm \,ae=\pm 2\sqrt{3}\] \[\Rightarrow \] \[{{a}^{2}}{{e}^{2}}=12\] Equation of eccentricity of second conic (hyperbola) \[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)\] \[\Rightarrow \] \[{{b}^{2}}={{a}^{2}}{{c}^{2}}-{{a}^{2}}\]s \[\Rightarrow \] \[9=12-{{a}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}=3\Rightarrow a=\sqrt{3}\]
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