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

Toolbox:
• Angle between the tangents of two curves is given by $\theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
• If $y=f(x),$ then $\bigg( \large\frac{dy}{dx} \bigg)_p=$ slope of the tangent to $y=f(x)$ at point $p$
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
$\tan\: \theta = \bigg| \large\frac{2\sqrt 2-(-2\sqrt 2)}{1+2\sqrt 2 \times -2\sqrt 2} \bigg|$
$= \bigg| \large\frac{2\sqrt 2+2\sqrt 2}{1-8} \bigg|$
$\tan\: \theta = \large\frac{4\sqrt 2}{7}$
$\therefore \theta = \tan^{-1} \bigg| \large\frac{4\sqrt 2}{7} \bigg|$