A) \[\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{16}=1\]
B) \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{12}=1\]
C) \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{8}=1\]
D) none of these
Correct Answer: C
Solution :
Let the ellipse be\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. It is given that \[e=\frac{1}{2}\] and\[ae=2\] Therefore,\[a=4\] Now, \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[{{b}^{2}}=12\] Thus, the required ellipse is\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{12}=1\]
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