The point on the curve y^{2}=x,the tangent at | vedclass

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(A) Given curve is $y^{2}=x$. Differentiating both sides with respect to $x$,we get: $2y \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2y}$. The slope

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given curve is $y^{2}=x$. Differentiating both sides with respect to $x$,we get: $2y \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2y}$. The slope of the tangent is given by $m = 	an(	heta)$. Given $	heta = 45^{\circ}$,so $m = 	an(45^{\circ}) = 1$. Equating the slope,we have: $\frac{1}{2y} = 1 \Rightarrow y = \frac{1}{2}$. Substituting $y = \frac{1}{2}$ into the curve equation $y^{2} = x$: $x = \left(\frac{1}{2}ight)^{2} = \frac{1}{4}$. Thus,the required point is $\left(\frac{1}{4}, \frac{1}{2}ight)$.

The point on the curve $y^{2}=x$,the tangent at which makes an angle $45^{\circ}$ with $X$-axis is

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