The line which is parallel to x -axis and crosses the curve y = x   at an angle 45 ° , is

The equation of a given curve is

$y=\sqrt{x}$ ...$\left(\mathrm{i}\right)$

On diffrenciating the above equation with respect to $x$ we get the slope

$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}$

The slope of the line at $\left(x_1, y_1\right), m_1=\frac{1}{2{\sqrt{x}}_1}$

and let line parallel to $x$-axis is $y=k$ ...$\left(\mathrm{ii}\right)$

Whose slope, $m_2=0$

Since, $45°$ is the angle between the line and the curve.

$\therefore \tan 45°=\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow 1=\left|\frac{\frac{1}{2\sqrt{x_1}}-0}{1}\right|$

$\Rightarrow x_1=\frac{1}{4}$

$\therefore y_1=\frac{1}{2}$ [from Eq.$\left(\mathrm{i}\right)$]
$\therefore$ The required equation is $y=\frac{1}{2}=k$  [from Eq.$\left(\mathrm{ii}\right)$]