Question

The angle of intersection of the curves $y=x^2$ and $x=y^2$ at $(1,1)$ is

• A. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(1\right)$
• C. ${90}^{°}$
• D. ${\tan }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is D

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

Explanation for the correct option:

Given equations of curves are $y=x^2$ and $x=y^2$

First, we have to find the point of intersection by equating both equations

solving the given equation

$$\begin{gathered} y=x^2 \\ \Rightarrow \sqrt{x}=x^2\left[\because y^2=x\right] \\ \Rightarrow x^4=x\left[\text{squaring on both sides}\right] \\ \Rightarrow x(x^3-1)=0 \\ \Rightarrow x=0,x=1 \end{gathered}$$

Therefore the points of intersection are $(0,0)\text{and}(1,1)$

Now, differentiating $y=x^2$ w.r.t '$x$'

$\Rightarrow \frac{dy}{dx}=2x$

At $(1,1)$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=2 \\ \Rightarrow m_1=2 \end{gathered}$$

Now, differentiating $x=y^2$ w.r.t '$x$'

$\Rightarrow x=2y\frac{dy}{dx}$

At $(1,1)$

$$\begin{gathered} \Rightarrow 2\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2} \\ \Rightarrow m_2=\frac{1}{2} \end{gathered}$$

The angle between two curves is

$$\begin{gathered} \tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right| \\ tan\theta =\left|\frac{{\displaystyle \frac{1}{2}}-2}{1+{\displaystyle \frac{1}{2}}\times 2}\right| \\ \tan \theta =\frac{3}{4} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$

Hence, option (D) is the correct answer.