Question

Find the angle between the curves $xy=2$ and $x^2+4y=0$.

Answer 1

Solving $xy=2$ and $x^2+4y=0$

$y=\frac{-x^2}{4}$

But $xy=2$

$\begin{array}{c}\Rightarrow x\left(\frac{-x^2}{4}\right)=2\\ \Rightarrow x^3=-8\\ \Rightarrow x=-2\end{array}$

$\Rightarrow y\left(\frac{-x^2}{4}\right)=\frac{-2^2}{4}=-1$

point of intersection is $p(-2,-1)$

$xy=2$ or $y=\frac{2}{x}$

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

$m_1={\left(\frac{dy}{dx}\right)}_p=\frac{-2}{4}=\frac{-1}{2}$

$\Rightarrow x^2+4y=0$

$\Rightarrow y=\frac{-x^2}{4}$

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{4}=\frac{-x}{2}$

$m_2={\left(\frac{dy}{dx}\right)}_p=\frac{-2}{-2}=1$

Let $\varphi$ be the angle between the given curves

$\tan \varphi \left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{-1}{2}-1}{1+\left(-\frac{1}{2}\right)1}\right|=\left|\frac{\frac{-3}{2}}{\frac{1}{2}}\right|=3$