Question

Angle between the given curves at there intersection points =>$y=4-x^2$ and $y=x^2$ be $\pm {\tan }^{-1}\frac{k\sqrt{m}}{n}.$.FInd $n-k-m$ ?

Answer 1

Given equation of curves
$y=4-x^2$ ....(1)
$\frac{dy}{dx}=-2x$
$y=x^2$ ....(2)
$\frac{dy}{dx}=2x$
Solving eqns (1) and (2), we get the point of intersection of the two curves as $(\sqrt{2},2)$ and $(-\sqrt{2},2)$
Slope of tangent to the curve given by eqn(1) is
$m_1={\left(\frac{dy}{dx}\right)}_{(-\sqrt{2},2)}=2\sqrt{2}$
$m_2={\left(\frac{dy}{dx}\right)}_{(\sqrt{2},2)}=-2\sqrt{2}$
Slope of tangent to the curve given by eqn(2) is
$m_3={\left(\frac{dy}{dx}\right)}_{(-\sqrt{2},2)}=-2\sqrt{2}$
$m_4={\left(\frac{dy}{dx}\right)}_{(\sqrt{2},2)}=2\sqrt{2}$
Angle between the curves at $(-\sqrt{2},2)$ is
$\tan \theta =\frac{m_1-m_3}{1+m_1{m}_3}$
$\Rightarrow \tan \theta =\frac{4\sqrt{2}}{-7}$
$\Rightarrow \theta =-{\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$
Angle between the curves at $(\sqrt{2},2)$ is
$\tan \theta =\frac{m_2-m_4}{1+m_2{m}_4}$
$\Rightarrow \tan \theta =\frac{-4\sqrt{2}}{-7}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$
$\therefore n-k-m=7-4-2=1$