A) \[\frac{{{(x-2)}^{2}}}{25}+\frac{{{(y-1)}^{2}}}{9}=1\]
B) \[\frac{{{(x-2)}^{2}}}{25}+\frac{{{(y-1)}^{2}}}{16}=1\]
C) \[\frac{{{(x-2)}^{2}}}{16}+\frac{{{(y-1)}^{2}}}{9}=1(d)\]
D) \[\frac{{{(x-2)}^{2}}}{9}+\frac{{{(y-1)}^{2}}}{25}=1\]
Correct Answer: D
Solution :
We know that the locus of a point so that the sum of its distances from two fixed points is constant, is an ellipse. Therefore, locus of the required point is an ellipse. Since the line joining the points\[(2,-3)\]and (2, 5) is parallel to y-axis. therefore axis of ellipse is vertical. \[\therefore \] \[2b=10\Rightarrow b=5\] and \[2be=8\] \[\Rightarrow \] \[e=\frac{8}{2\times 5}=\frac{4}{5}\] \[\therefore \] \[{{a}^{2}}={{b}^{2}}(1-{{e}^{2}})\] \[{{a}^{2}}=25\left( \frac{9}{25} \right)\Rightarrow a=3\] Mid-point of\[(2,-3)\]and (2,5) is (2, 1) which is the centre of the ellipse. \[\therefore \]Equation of required ellipse is \[\frac{{{(x-2)}^{2}}}{9}+\frac{{{(y-1)}^{2}}}{25}=1\]
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