A) \[\frac{{{(x+3)}^{2}}}{24}+\frac{{{(y+2)}^{2}}}{25}=1\]
B) \[\frac{{{(x-3)}^{2}}}{24}+\frac{{{(y-2)}^{2}}}{25}=1\]
C) \[\frac{{{(x+3)}^{2}}}{24}+\frac{{{(y+2)}^{2}}}{24}=1\]
D) \[\frac{{{(x-3)}^{2}}}{25}+\frac{{{(y-2)}^{2}}}{24}=1\]
Correct Answer: D
Solution :
Given, foci of an ellipse are (2, 2) and (4, 2) and major axis is of length 10. \[\therefore \] \[2ae=2\] and \[2a=10\Rightarrow a=5\] \[\therefore \] \[2\times 5\times e=2\] \[\Rightarrow \]\[e=\frac{1}{5}\] \[\because \] \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\Rightarrow {{b}^{2}}=25\left( 1-\frac{1}{25} \right)\] \[\Rightarrow \] \[{{b}^{2}}=24\] Centre of ellipse\[=\]Mid-point of foci \[=\left( \frac{2+4}{2},\frac{2+2}{2} \right)=(3,2)\] Hence, equation of ellipse is \[\frac{{{(x-3)}^{2}}}{25}+\frac{{{(y-2)}^{2}}}{24}=1\]
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