Question

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

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

Answer 1

The correct option is D $ta{n}^{-1}\left(\frac{3}{4}\right)$

$y=x^2\Rightarrow \frac{dy}{dx}=m_1=2x$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,1)}=2=m_1$ and $x=y^2\Rightarrow 1=2y\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=m_2=\frac{1}{2y}\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,1)}=\frac{1}{2}$
$\therefore$ Angle of intersection, $tan\theta =\frac{m_1-m_2}{1+m_1{m}_2}=\frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}}=\frac{3}{4}$
$\Rightarrow \theta =ta{n}^{-1}\left(\frac{3}{4}\right)$.