Find the angle of intersection of the curves $y = 4 - {x^2}$ and $y = {x^2}$.

We have, $y = 4 - {x^2}$ …..(i)

and $y = {x^2}$ ….(ii)

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

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

$\Rightarrow$ ${m_1} = - 2x$

and ${m_2} = 2x$

From Eqs. (i) and (ii), ${x^2} = 4 - {x^2}$

$\Rightarrow$ $2{x^2} = 4$

$\Rightarrow$ ${x^2} = 2$

$\Rightarrow$ $x = \pm \sqrt 2$

$y = {x^2} = {( \pm \sqrt 2 )^2} = 2$

Therefore, the points of intersection are $(\sqrt 2 ,2)$ and $( - \sqrt 2 ,2)$.

For point $( + \sqrt 2 ,2),$ ${m_1} = - 2x = - 2 \cdot \sqrt 2 = - 2\sqrt 2$

and ${m_2} = 2x = 2\sqrt 2$

and for point $(\sqrt 2 ,2),$ $\tan \theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| = \left| {\frac{{ - 2\sqrt 2 - 2\sqrt 2 }}{{1 - 2\sqrt 2 \cdot 2\sqrt 2 }}} \right| = \left| {\frac{{ - 4\sqrt 2 }}{{ - 7}}} \right|$

Therefore,$\theta = {\tan ^{ - 1}}\left( {\frac{{4\sqrt 2 }}{7}} \right)$