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

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(D) Let the line be $y = c$. The curve is $y = \sqrt{x}$.At the point of intersection $P(x, y)$,we have $y = \sqrt{x}$,so $x = y^2$. Since $y = c$,the

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

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(D) Let the line be $y = c$. The curve is $y = \sqrt{x}$.At the point of intersection $P(x, y)$,we have $y = \sqrt{x}$,so $x = y^2$. Since $y = c$,the point $P$ is $(c^2, c)$.The slope of the tangent to the curve $y = \sqrt{x}$ at any point is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.At point $P(c^2, c)$,the slope is $m = \frac{1}{2\sqrt{c^2}} = \frac{1}{2c}$.The angle between the curve and the line $y = c$ (which is parallel to the $x$-axis) is given as $\frac{\pi}{4}$.The slope of the line $y = c$ is $0$. The angle $	heta$ between the curve and the line is given by $	an(	heta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Here,$	an\left(\frac{\pi}{4}ight) = \left| \frac{\frac{1}{2c} - 0}{1 + \frac{1}{2c} \cdot 0} ight| = \frac{1}{2c}$.Since $1 = \frac{1}{2c}$,we get $2c = 1$,which implies $c = \frac{1}{2}$.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 $\frac{\pi}{4}$ is:

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