A) \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\]
B) \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\]
C) \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]
D) \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{5}=1\]
Correct Answer: A
Solution :
Given, vertices \[(\pm \,a,\,0)=(\pm \,5,0)\] and foci \[(\pm \,\,4,\,\,0)=(\pm \,\,ae,\,0)\] \[\therefore \] \[a=5\] and \[ae=4\] \[\Rightarrow \] \[e=\frac{4}{a}=\frac{4}{5}\] \[\because \] \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[\therefore \] \[{{b}^{2}}=25\left( 1-\frac{16}{25} \right)\] \[\Rightarrow \] \[{{b}^{2}}=9\] \[\because \] \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\therefore \] \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\]
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