The point on the curve y^{2}=x where the tang | vedclass

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(B) Given the curve equation: $y^{2} = x$  $(1)$Slope of the tangent $m$ is given by $\frac{dy}{dx}$.We know that the slope $m = 	an(	heta)$,where $

Vedclass · Accepted Answer

$(\frac{1}{4}, \frac{1}{2})$

Vedclass · Answer

(B) Given the curve equation: $y^{2} = x$  $(1)$Slope of the tangent $m$ is given by $\frac{dy}{dx}$.We know that the slope $m = 	an(	heta)$,where $	heta$ is the angle with the $x$-axis.Here,$	heta = \frac{\pi}{4}$,so $m = 	an(\frac{\pi}{4}) = 1$.Differentiating equation $(1)$ with respect to $x$:$2y \frac{dy}{dx} = 1$Substituting $m = \frac{dy}{dx} = 1$:$2y(1) = 1 \Rightarrow y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into equation $(1)$:$(\frac{1}{2})^{2} = x \Rightarrow x = \frac{1}{4}$.Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

The point on the curve $y^{2}=x$ where the tangent makes an angle of $\pi / 4$ with the $x$-axis is

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