Question

The angle between tangents to the curves $y=x^2$ and $x=y^2$ at $\left(1,1\right)$ is :

• A. $0$
• B. ${\tan }^{-1}1$
• C. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is C

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

Explanation for correct option

Find the angle between tangents to the curves

Given, the curves are

$x=y^2......\left(1\right)$

$y=x^2......\left(2\right)$

Differentiating equations$\left(1\right)$ with respect to $\mathrm{x}$,

$$\begin{gathered} \Rightarrow 2y\frac{dy}{dx}=1 \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2y} \\ \therefore m_1=\frac{1}{2}\left[\text{At}\left(1,1\right)\right]......\left(3\right) \end{gathered}$$

Differentiate the Equations $\left(2\right)$ with respect to $x$,

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=2x \\ \therefore m_1=2\left[\text{At}\left(1,1\right)\right]......\left(4\right) \end{gathered}$$

We know that, the angle between the tangents is given by:

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

Substituting the value of $m_1$ and $m_2$ in the above formula,

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

Hence, the correct answer is Option (C).