A) 1
B) 7
C) 5
D) 9
Correct Answer: B
Solution :
We have, \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ... (i) and \[\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\] ?. (ii) for hyperbola, \[{{e}^{2}}=1+\frac{{{b}^{2}}}{{{a}^{2}}}=1+\frac{81}{144}=\frac{225}{144}\] \[\therefore \] \[e=\frac{15}{12}=\frac{5}{4}\] \[i.e.,\,e>1\] also, \[{{a}^{2}}=\frac{144}{25}\] Hence, the foci are \[(\pm \,ae,\,0)\] i.e.,\[\left( \pm \frac{12}{5}.\frac{5}{4},0 \right)=(\pm \,3,\,0)\] Now, the foci coincide \[\therefore \] for ellipse \[ae=3\]or \[{{a}^{2}}{{e}^{2}}=9\] or \[{{a}^{2}}\left( 1-\frac{{{b}^{2}}}{{{a}^{2}}} \right)=9\] \[\Rightarrow \] \[{{a}^{2}}-{{b}^{2}}=9\] or \[16-9={{b}^{2}}\]\[16-9={{b}^{2}}\] \[\Rightarrow \] \[{{b}^{2}}=7\]
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