At which point on the curve y^2 = x does the | vedclass

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(D) Given the curve is $y^2 = x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,which implies $\frac{dy}{dx} = \frac{1}{2y}$. The

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$(1/4, 1/2)$

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(D) Given the curve is $y^2 = x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,which implies $\frac{dy}{dx} = \frac{1}{2y}$. The slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{1}{2y_1}$. Since the tangent makes an angle of $45^{\circ}$ with the $x-$axis,the slope $m = 	an(45^{\circ}) = 1$. Equating the slopes,$\frac{1}{2y_1} = 1$,which gives $y_1 = \frac{1}{2}$. Substituting $y_1 = \frac{1}{2}$ into the curve equation $y^2 = x$,we get $x_1 = (\frac{1}{2})^2 = \frac{1}{4}$. Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

At which point on the curve $y^2 = x$ does the tangent make an angle of $45^{\circ}$ with the $x-$axis?

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