A) \[\frac{{{x}^{2}}}{81}+\frac{{{y}^{2}}}{45}=1\]
B) \[\frac{4{{x}^{2}}}{81}+\frac{4{{y}^{2}}}{45}=1\]
C) \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{5}=1\]
D) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]
Correct Answer: B
Solution :
\[{{\left( \frac{2}{3} \right)}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\] and \[\frac{2{{b}^{2}}}{a}=5\] Þ \[a=\frac{81}{4}\], \[b=\frac{45}{4}\] Hence the equation is\[\frac{4{{x}^{2}}}{81}+\frac{4{{y}^{2}}}{45}=1\].
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