Step 1

$y=4-x^2\: \: \: \: (1)$

$y=x^2\: \: \: \: \: \: \: \:\:\: \: (2)$

On solving the equation we get

$x^2=4-x^2 \Rightarrow 2x^2=4$

$ x^2=2\: x= \pm \sqrt 2$

$ \Rightarrow y= 2$

Consider $y=4-x^2$

differentiating w.r.t $x$ we get,

$ \bigg( \large\frac{dy}{dx} \bigg)=-2x\: \therefore \bigg(\large\frac{dy}{dx} \bigg)_{(\sqrt 2, 2)}=-2\sqrt 2$

Let $m_1=-2\sqrt 2$

Consider $y=x^2$

differentiating w.r.t $x$ we get,

$ \large\frac{dy}{dx}=2x\: \therefore \bigg( \large\frac{dy}{dx} \bigg)_{(\sqrt 2, 2)}=2\sqrt 2$

Let this be $m_2$

The angle between the two curves is given as

$ \tan \: \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$

Substituting for $m_1\: and \: m_2$ we get

$ \tan\: \theta = \bigg| \large\frac{2\sqrt 2-(-2\sqrt 2)}{1+2\sqrt 2 \times -2\sqrt 2} \bigg|$

$ = \bigg| \large\frac{2\sqrt 2+2\sqrt 2}{1-8} \bigg|$

$ \tan\: \theta = \large\frac{4\sqrt 2}{7}$

$ \therefore \theta = \tan^{-1} \bigg| \large\frac{4\sqrt 2}{7} \bigg|$