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 :
Since given that foci of an ellipse are (2, 2) and (4,2). \[\therefore \]Focal distance = 2 and centre of ellipse is (3, 2) ...(i) \[\Rightarrow \] \[2ae=2\] and \[2a=10\] (given) \[\Rightarrow \] \[a=5\] ...(ii) \[\Rightarrow \] \[e=\frac{1}{5}\] ...(iii) Also, we know \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[\therefore \] \[{{b}^{2}}=25\left( 1-\frac{1}{25} \right)=\frac{24}{25}\times 25=24\] \[\Rightarrow \] \[{{b}^{2}}=24\] ...(iv) From Eqs. (i), (ii), (hi) and (iv) Equation of ellipse is \[\frac{{{(x-3)}^{2}}}{25}+\frac{{{(y-2)}^{2}}}{24}=1\]
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