STD 11 - 10.2 parabola,ellipse,hyperbola — JEE कक्षा 11 साइन्स — Question

$y = 2{x^2} + x$

$ \Rightarrow \,{x^2} + \frac{x}{2} = \frac{y}{2}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{y}{2} + \frac{1}{{16}}$

$ \Rightarrow \,{\left( {x + \frac{1}{4}} \right)^2} = \frac{1}{2}\left( {y + \frac{1}{8}} \right)$

${X^2} = \frac{1}{2}Y$.....$(i)$

यहाँ $A = \frac{1}{8}$ $X = x + \frac{1}{4},\,Y = y + \frac{1}{8}$, $(i)$ की नाभि $\left( {0,\frac{1}{8}} \right)$ है

अर्थात् $X = 0$, $Y = \frac{1}{8}$

 $x + \frac{1}{4} = 0$, $y + \frac{1}{8} = \frac{1}{8}$

$ \Rightarrow \,x =  - \frac{1}{4},$ $y = 0$

अत: परवलय की नाभि $\left( { - \frac{1}{4},\,0} \right)$ है।