Question

The straight line which is parallel to x axis and crosses the curve y $=\sqrt{x}$ at an angle of ${45}^0$ is

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

Answer 1

Answer: (C)

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

The required line is parallel to the $X$-axis.
Let the equation of the line be $y-c=0-\left(1\right)$ ,where ${}^{\prime }{c}^{\prime }$ is a constant.
Line $\left(1\right)$ intersects the given curve at $(c^2,c)$ at an angle of $\frac{\pi }{4}$.
Slope of the tangent to the curve at this point is $\frac{1}{2c}$. ( since $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$, when $y=c$, $\sqrt{x}=c$)
Angle between this tangent and the given line is $\frac{\pi }{4}$.
Let $m_1=\frac{1}{2c}$ and $m_2=0$.
Applying the formula for the angle between two lines,
$\frac{\pi }{4}={\tan }^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|\Rightarrow 1=\left|\frac{\left(1/2c\right)-0}{1}\right|\Rightarrow c=\pm \frac{1}{2}$
$\therefore$The equation of the line is $y=\frac{1}{2}$. (since for $y=\sqrt{x}$, y can not be negative)