Conic Sections Question 480
Question: The co-ordinates of the foci of the ellipse $ 3x^{2}+4y^{2}-12x-8y+4=0 $ are
Options:
A) (1, 2), (3, 4)
B) (1, 4), (3, 1)
C) (1, 1), (3, 1)
D) (2, 3), (5, 4)
Show Answer
Answer:
Correct Answer: C
Solution:
$ 3x^{2}-12x+4y^{2}-8y=-4 $
therefore $ 3{{(x-2)}^{2}}+4{{(y-1)}^{2}}=12 $
therefore $ \frac{{{(x-2)}^{2}}}{4}+\frac{{{(y-1)}^{2}}}{3}=1 $
therefore $ \frac{X^{2}}{4}+\frac{Y^{2}}{3}=1 $
$ e=\sqrt{1-\frac{3}{4}}=\frac{1}{2} $ .Foci are $ ( X=\pm 2\times \frac{1}{2},Y=0 ) $
i.e., $ (x-2=\pm 1,y-1=0) $
$ =(3,1) $ and $ (1,1) $ .
BETA
