The line which is parallel to the x-axis and | vedclass

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Question

(C) Let the equation of the line parallel to the $x$-axis be $y = \lambda$. Since the line intersects the curve $y = \sqrt{x}$,we have $\lambda = \sqr

Vedclass · Accepted Answer

$y = \frac{1}{2}$

Vedclass · Answer

(C) Let the equation of the line parallel to the $x$-axis be $y = \lambda$. Since the line intersects the curve $y = \sqrt{x}$,we have $\lambda = \sqrt{x}$,which implies $x = \lambda^2$. Thus,the point of intersection is $P(\lambda^2, \lambda)$. The slope of the curve $y = \sqrt{x}$ is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$. At the point $P(\lambda^2, \lambda)$,the slope $m$ is $\frac{1}{2\sqrt{\lambda^2}} = \frac{1}{2|\lambda|}$. The angle between the line (slope $0$) and the curve (slope $m$) is $45^\circ$. Therefore,$	an(45^\circ) = \left| \frac{m - 0}{1 + m \cdot 0} ight| = |m|$. $1 = \left| \frac{1}{2\lambda} ight|$,which gives $2|\lambda| = 1$,so $\lambda = \pm \frac{1}{2}$. Since $y = \sqrt{x}$ implies $y \ge 0$,we take $\lambda = \frac{1}{2}$. Thus,the equation of the line is $y = \frac{1}{2}$.

The line which is parallel to the $x$-axis and intersects the curve $y = \sqrt{x}$ at an angle of $45^\circ$ is given by:

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