Find the angle of intersection of the following curves: `x y=6` and `x^2y=12`
`y^2=4x` and`x^2=4y`

Answer

`A= tan ^{-1} (frac{3}{11})`

Explanation

\begin{aligned}
&x y=6 \text { and } x^{2} y=12 \\
&\text { Hence } \frac{x^{2} y}{x y} \Rightarrow x=2 \Rightarrow y=3 \\
&x y=6 \Rightarrow y=\frac{6}{x} \Rightarrow \frac{d y}{d x}=\frac{-6}{x^{2}}=\frac{-3}{2}
\end{aligned}
$$
Let `m_1=\frac{-3}{2}`

And `x^{2} y=1`
...