Conic Sections Question 31
Question: The focus of the parabola $ y=2x^{2}+x $ is
[MP PET 2000]
Options:
A) (0, 0)
B) $ ( \frac{1}{2},\ \frac{1}{4} ) $
C) $ ( -\frac{1}{4},\ 0 ) $
D) $ ( -\frac{1}{4},\ \frac{1}{8} ) $
Show Answer
Answer:
Correct Answer: C
Solution:
The given equation of parabola is $ y=2x^{2}+x $
$ \Rightarrow x^{2}+\frac{x}{2}=\frac{y}{2} $
$ \Rightarrow {{( x+\frac{1}{4} )}^{2}}=\frac{y}{2}+\frac{1}{16} $
$ \Rightarrow {{( x+\frac{1}{4} )}^{2}}=\frac{1}{2}( y+\frac{1}{8} ) $
It can be written as, $ X^{2}=\frac{1}{2}Y $ …..(i)
Here $ A=\frac{1}{8} $ , focus of (i) is $ ( 0,\frac{1}{8} ) $ i.e. $ X=0 $ , $ Y=\frac{1}{8} $
therefore $ x+\frac{1}{4}=0 $ , $ y+\frac{1}{8}=\frac{1}{8} $
$ \Rightarrow x=-\frac{1}{4}, $
$ y=0 $
i.e. focus of given parabola is $ ( -\frac{1}{4},0 ) $ .
BETA
