Question

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

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

Answer 1

Answer: (C)

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

Answer 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 $\left( k ^{2}, k \right)$
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 $\left( k ^{2}, k \right)$ is
$\tan \left(+\frac{\pi}{4}\right)=\pm 1$
$\Rightarrow\left(\frac{ dy }{ dx }\right)_{\left( k ^{2}, k \right)}=\pm 1$
$\Rightarrow\left(\frac{1}{2 \sqrt{ x }}\right)_{\left( k ^{2}, k \right)}=\pm 1$
$\Rightarrow k =\pm \frac{1}{2}$
Thus $y =\pm \frac{1}{2}$