A) \[5{{x}^{2}}+3{{y}^{2}}-32=0\]
B) \[3{{x}^{2}}+5{{y}^{2}}-32=0\]
C) \[5{{x}^{2}}+3{{y}^{2}}-48=0\]
D) \[3{{x}^{2}}+5{{y}^{2}}-15=0\]
Correct Answer: B
Solution :
Let the equation of the ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Which will pass through (? 3, 1) if \[\frac{9}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=1\] and eccentricity \[=e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\frac{2}{5}}\] \[\Rightarrow \frac{2}{5}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{3}{5}\] \[\Rightarrow {{b}^{2}}=\frac{3}{5}{{a}^{2}}\] Thus \[\frac{9}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=1\] \[\square \,\,\,\,\,\,\,\,\frac{9}{{{a}^{2}}}+\frac{5}{3{{a}^{2}}}=1\] \[\Rightarrow 27+5=3{{a}^{2}}=32\] \[\Rightarrow {{a}^{2}}=\frac{32}{3},{{b}^{2}}=\frac{3}{5}\times \frac{32}{3}=\frac{32}{5}\] Required equation of the ellipse is \[3{{x}^{2}}+5{{y}^{2}}=32\]
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