A) \[(0,0)\]
B) \[\left( \frac{1}{2},\frac{1}{4} \right)\]
C) \[\left( -\frac{1}{4},0 \right)\]
D) \[\left( -\frac{1}{4},\frac{1}{8} \right)\]
Correct Answer: C
Solution :
The given equation of parabola is \[y=2{{x}^{2}}+x\] \[\Rightarrow \] \[{{x}^{2}}+\frac{x}{2}=\frac{y}{2}\] \[\Rightarrow \] \[{{x}^{2}}+\frac{x}{2}+\frac{1}{16}=\frac{y}{2}+\frac{1}{16}\] \[\Rightarrow \] \[{{\left( x+\frac{1}{4} \right)}^{2}}=\frac{1}{2}\left( y+\frac{1}{8} \right)\] It can be rewritten as \[{{X}^{2}}=\frac{1}{2}Y\] ... (i) where \[x+\frac{1}{4}=X\] and \[y+\frac{1}{8}=Y\] On comparing with \[{{X}^{2}}=4\text{ }AY,\]we get \[A=\frac{1}{8}\] focus of Eq.(i) is \[\left( 0,\frac{1}{8} \right)\] ie, \[X=0,\] \[Y=\frac{1}{8}\] \[\Rightarrow \] \[x+\frac{1}{4}=0,\]\[y+\frac{1}{8}=\frac{1}{8}\] \[\Rightarrow \]\[x=-\frac{1}{4},\] \[y=0\] \[\therefore \]Focus of given parabola is \[\left( -\frac{1}{4},0 \right)\].
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