Question

The line which is parallel to the x-axis and crosses the curve $y=\sqrt{x}$ at an angle of $45{}^\circ $ is equal to
A. $x=\dfrac{1}{4}$
B. $y=\dfrac{1}{4}$
C. $y=\dfrac{1}{2}$
D. $y=1$

Answer 1

Hint: In this question, we are to find the equation of the line that is parallel to the x-axis and crosses a curve at an angle. In order to find this, we need to consider the line parallel to the x-axis as $y=k$. By using this, we can frame the required equation.

Formula used: The equation of the x-axis is $y=0$.
The slope of a line is $m=\tan \theta $
The slope of a curve is
$m={{\left( \dfrac{dy}{dx} \right)}_{P(x,y)}}$

Complete step by step solution: Consider the equation of the line that is parallel to the x-axis is
$y=k\text{ }...(1)$
It is given that equation (1) crosses the curve $y=\sqrt{x}\text{ }...(2)$
So, their point of intersection is calculated by substituting (1) in (2)
$\begin{align}
  & y=\sqrt{x} \\
 & \Rightarrow \sqrt{x}=k \\
 & \Rightarrow x={{k}^{2}} \\
\end{align}$
Thus, the point of intersection is $({{k}^{2}},k)$
The slope of (2) is calculated by
$m={{\left( \dfrac{dy}{dx} \right)}_{P(x,y)}}$
(Since it is a curve, we use derivation)
$\begin{align}
  & m={{\left( \dfrac{d}{dx}\sqrt{x} \right)}_{({{k}^{2}},k)}} \\
 & \text{ }={{\left( \dfrac{1}{2\sqrt{x}} \right)}_{_{({{k}^{2}},k)}}} \\
 & \text{ }=\dfrac{1}{2\sqrt{{{k}^{2}}}} \\
 & \text{ }=\dfrac{1}{2k} \\
\end{align}$
Since it is given these two intersects at an angle of $45{}^\circ $, we get
  $\begin{align}
  & m=\tan \theta \\
 & \text{ }=\tan 45{}^\circ \\
 & \text{ }=1 \\
\end{align}$
Then, we can write
$\begin{align}
  & 1=\dfrac{1}{2k} \\
 & \Rightarrow k=\dfrac{1}{2} \\
\end{align}$
Therefore, the equation is $y=\dfrac{1}{2}$

Thus, Option (C) is correct.

Note: Here we need to remember that the line parallel to the x-axis is $y=k$. Since the line crosses and intersects a curve, the derivative function at the point of intersection gives the slope of the curve. With all these, we are able to frame the equation of the line.