The points on the curve y^2 = x + sin x at wh | vedclass

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(C) Given curve is $y^2 = x + \sin x$ ...$(i)$By differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1 + \cos x$,which implies $\frac{dy}{d

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a parabola

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(C) Given curve is $y^2 = x + \sin x$ ...$(i)$By differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1 + \cos x$,which implies $\frac{dy}{dx} = \frac{1 + \cos x}{2y}$.The slope of the normal is $m = -\frac{1}{dy/dx} = -\frac{2y}{1 + \cos x}$.For the normal to be parallel to the $Y$-axis,the slope must be undefined,i.e.,$1 + \cos x = 0$.This implies $\cos x = -1$,which means $x = (2n+1)\pi$ for some integer $n$.At these points,$\sin x = 0$.Substituting $\sin x = 0$ into the original equation $(i)$,we get $y^2 = x + 0$,or $y^2 = x$.This is the equation of a parabola.

The points on the curve $y^2 = x + \sin x$ at which the normal is parallel to the $Y$-axis lie on

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