A) \[4\sqrt{5}\]
B) \[\frac{49}{4}{{x}^{2}}-\frac{51}{196}{{y}^{2}}=1\]
C) 3
D) 4
Correct Answer: D
Solution :
Given equation may be written as \[\frac{{{x}^{2}}}{5}+\frac{{{y}^{2}}}{9}=1.\] Comparing the given equation with standard equation, we get \[{{a}^{2}}=5\] and \[{{b}^{2}}=9.\] We also know that in an ellipse (where \[{{b}^{2}}>{{a}^{2}})\] \[{{a}^{2}}={{b}^{2}}(1-{{e}^{2}})\,\lambda \] or \[{{e}^{2}}=\frac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}}=\frac{9-5}{9}=\frac{4}{9}\] or \[e=\frac{2}{3}.\] Therefore distance between foci \[=2be=2\times 3\times \frac{2}{3}=4\].
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