The point on the curve y2 = x where tangent makes $45^\circ$ angle with x-axis is ____________________ .

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KollegeApply · Accepted Answer

($\frac{1}{4}$, $\frac{1}{2}$) Let the required point be (x1, y1). The tangent makes an angle of 45o with the x-axis. $\therefore$ Slope of the tangent = tan 45o = 1 \text { Since, the point lies on the curve } . \text { Hence }, {y_1}^2 = x_1 \text { Now,} y^2 = x \Rightarrow 2ydy/dx = 1 \Rightarrow dy/dx = $\frac{1}{2}$y \text { Slope of the tangent } = \left( dy/dx \right)_\left( x_1 , y_1 \right) =$\frac{1}{2}$ y_1 \text { Given }: $\frac{1}{2}$ y_1 = 1 \Rightarrow 2 y_1 = 1 \Rightarrow y_1 = $\frac{1}{2}$ \text { Now,} x_1 = {y_1}^2 = \left( $\frac{1}{2}$ \right)^2 = $\frac{1}{4}$ \therefore \left( x_1 , y_1 \right) = \left( $\frac{1}{4}$, $\frac{1}{2}$ \right)

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