
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
164.7k+ views
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.
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.
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