Conic Sections Question 334
Question: The eccentricity of the curve represented by the equation $ x^{2}+2y^{2}-2x+3y+2=0 $ is
[Roorkee 1998]
Options:
A) 0
B) 1/2
C) $ 1/\sqrt{2} $
D) $ \sqrt{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Equation $ x^{2}+2y^{2}-2x+3y+2=0 $ can be written as $ \frac{{{(x-1)}^{2}}}{2}+{{( y+\frac{3}{4} )}^{2}}=\frac{1}{16} $
therefore $ \frac{{{(x-1)}^{2}}}{(1/8)}+\frac{{{( y+\frac{3}{4} )}^{2}}}{(1/16)}=1 $ , which is an ellipse with $ a^{2}=\frac{1}{8} $ and $ b^{2}=\frac{1}{16} $
$ \therefore \frac{1}{16}=\frac{1}{8}(1-e^{2}) $
therefore $ e^{2}=1-\frac{1}{2} $
therefore $ e=\frac{1}{\sqrt{2}} $ .
BETA
