A) 8
B) 6
C) 50
D) 32
E) 10
Correct Answer: B
Solution :
Given equation of ellipse can be rewritten as \[\frac{{{(x-2)}^{2}}}{25}+\frac{{{(y+3)}^{2}}}{16}=1\] \[\Rightarrow \] \[\frac{{{X}^{2}}}{25}+\frac{{{Y}^{2}}}{16}=1\] Where \[X=x-2,Y=y+3\] Here, \[a>b\] \[\therefore \] \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\] \[\therefore \] Focus\[(\pm ae,0)=(\pm \text{ }3,0)\] \[\Rightarrow \] \[x-2=\pm \text{ }3,\text{ }y+3=0\] \[\Rightarrow \] \[x=5,-1,\text{ }y=-3\] \[\therefore \]Foci are\[(-1,-3)\]and\[(5,-3)\]. Distance between\[(2,-3)\]and\[(-1,-3)\] \[=\sqrt{{{(2+1)}^{2}}+{{(-3+3)}^{2}}}=3\] and distance between\[(2,-3)\]and\[(5,-3)\] \[=\sqrt{{{(2-5)}^{2}}+{{(-3+3)}^{2}}}=3\] Hence, sum of the distance of point\[(2,-3)\] from the foci\[=3+3=6\]
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