Question

The sum of the distances of a point $ (2,-3) $ from the foci of an ellipse $ 16{{(x-2)}^{2}}+25({{y}^{4}}- $ $ 3{{)}^{2}}=400 $ is

• A. 8
• B. 6
• C. 50
• D. 32
• E. 10

Answer 1

Answer: (B)

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 $