Question

The line which is parallel to $x$-axis and crosses the curve $y=\sqrt{x}$ at an angle of ${45}^o$, is

• A. $x=\frac{1}{4}$
• B. $y=\frac{1}{4}$
• C. $y=\frac{1}{2}$
• D. $y=1$

Answer 1

Answer: (C)

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

Given equation of a line parallel to $x$-axis is $y=k$
Given equation of the curve is $y=\sqrt{x}$
On solving equation of line with the equation of curve, we get $x=k^2$
Thus the intersecting point is $(k^2,k)$
It is given that the line $y=k$ intersect the curve $y=\sqrt{x}$ at an angle of $\frac{\pi }{4}$.

This means that the slope of the tangent to $y=\sqrt{x}$ at $(k^2,k)$ is $\tan (+\frac{\pi }{4})=\pm 1$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(k^2,k)}=\pm 1$

$\Rightarrow {\left(\frac{1}{2\sqrt{x}}\right)}_{(k^2,k)}=\pm 1$

$\Rightarrow k=\pm \frac{1}{2}$

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