Question

The angle between the curves $y=x^2$ and $y^2-x=0$at the point $\left(1,1\right)$ is

• A. $\frac{\mathrm{\pi }}{2}$
• B. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{\mathrm{\pi }}{4}$
• E. ${\tan }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is E

${\tan }^{-1}\left(\frac{3}{4}\right)$

Step 1: Find the slope of both the curves.

Differentiate the equation $y=x^2$with respect to $x$

$$\begin{gathered} \Rightarrow y=x^2 \\ \Rightarrow \frac{dy}{dx}=2x \\ \Rightarrow m_2=2\left[\text{At}\left(1,1\right)\right] \end{gathered}$$

Again, Differentiate the equation $y^2-x=0$with respect to $x$

$$\begin{gathered} \Rightarrow y^2-x=0 \\ {y}^2=x \\ \Rightarrow 2y\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2y} \\ \Rightarrow m_1=\frac{1}{2}\left[\text{At}\left(1,1\right)\right] \end{gathered}$$

Step 2: Find the angle between the curves.

We know that

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

Substitute the value of $m_1\text{and}{m}_2$in the above Equation

$$\begin{gathered} \tan \theta =\left|\frac{2-{\displaystyle \frac{1}{2}}}{1+\left({\displaystyle \frac{1}{2}}\right)2}\right| \\ \tan \theta =\left|\frac{{\displaystyle \frac{3}{2}}}{1+1}\right| \\ \tan \theta =\left|\frac{{\displaystyle \frac{3}{2}}}{2}\right| \\ \tan \theta =\frac{3}{2}\times \frac{1}{2} \\ \tan \theta =\frac{3}{4} \\ \theta ={\tan }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$

Hence, the correct answer is Option E.