Question

The angle between the curves $x^2+4y^2=32$ and $x^2-y^2=12$ is -

Answer 1

$\begin{array}{c}{x}^2+4y^2=\mathrm{32........}\left(1\right)\\ x^2-y^2=12\Rightarrow x^2=12-y^2\end{array}$
put value of $x^2$ in $\left(1\right)$
$\begin{array}{c}12+y^2+4y^2=32\\ y=\pm 4\\ \left(4,2\right)\\ 4x+8y=32\\ slope\Rightarrow \frac{-1}{2}=m_1\\ slope\Rightarrow 2=m_2\end{array}$
so we can see
$m_1{m}_2=\frac{-1}{2}2=-1$
tangents are perpendicular to each other