Conic Sections Question 537
Question: The foci of the ellipse $ 25{{(x+1)}^{2}}+9{{(y+2)}^{2}}=225 $ are at
[MNR 1991; MP PET 1998; UPSEAT 2000]
Options:
A) (-1, 2) and (-1, -6)
B) (-1, 2) and (6, 1)
C) (1, -2) and (1, -6)
D) (-1, -2) and (1, 6)
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{{{(x+1)}^{2}}}{\frac{225}{25}}+\frac{{{(y+2)}^{2}}}{\frac{225}{9}}=1 $
$ a=\sqrt{\frac{225}{25}}=\frac{15}{5},b=\sqrt{\frac{225}{9}}=\frac{15}{3} $
therefore $ e=\sqrt{1-\frac{9}{25}}=\frac{4}{5} $
Focus $ =( -1,-2\pm \frac{15}{3}.\frac{4}{5} )=(-1,-2\pm 4) $ =(-1,2); (-1,-6) .
BETA
