Prove that the curves

Given:

Curves $x y=4$   .....(1)

$\& x^{2}+y^{2}=8 \ldots(2)$

Solving $(1) \&(2)$, we get

$\Rightarrow x y=4$

$\Rightarrow x=\frac{4}{y}$

Substituting $x=\frac{4}{y}$ in $x^{2}+y^{2}=8$, we get,

$\Rightarrow\left(\frac{4}{y}\right)^{2}+y^{2}=8$

$\Rightarrow \frac{16}{y^{2}}+y^{2}=8$

$\Rightarrow 16+y^{4}=8 y^{2}$

$\Rightarrow y^{4}-8 y^{2}+16=0$

We will use factorization method to solve the above equation

$\Rightarrow y^{4}-4 y^{2}-4 y^{2}+16=0$

$\Rightarrow y^{2}\left(y^{2}-4\right)-4\left(y^{2}-4\right)=0$

$\Rightarrow\left(y^{2}-4\right)\left(y^{2}-4\right)=0$

$\Rightarrow y^{2}-4=0$

$\Rightarrow y^{2}=4$

$\Rightarrow y=\pm 2$

Substituting $y=\pm 2$ in $x=\frac{4}{y}$, we get,

$\Rightarrow x=\frac{4}{+2}$

$\Rightarrow x=\pm 2$

$\therefore$ The point of intersection of two curves $(2,2) \&$

$(-2,-2)$

First curve $x y=4$

$\Rightarrow 1 \times y+x \cdot \frac{d y}{d x}=0$

$\Rightarrow x \cdot \frac{d y}{d x}=-y$

$\Rightarrow m_{1}=\frac{-y}{x} \ldots(3)$

Second curve is $x^{2}+y^{2}=8$

Differentiating above w.r.t $x$,

$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}=0$

$\Rightarrow y \cdot \frac{d y}{d x}=-x$

$\Rightarrow m_{2}=\frac{d y}{d x}=\frac{-x}{y} \ldots(4)$

At $(2,2)$, we have,

$m_{1}=\frac{-y}{x}$

$\Rightarrow \frac{-2}{2}$

$m_{1}=-1$

At $(2,2)$, we have,

$\Rightarrow m_{2}=\frac{-x}{y}$

$\Rightarrow \frac{-2}{2}$

$\Rightarrow m_{2}=-1$

Clearly, $m_{1}=m_{2}=-1$ at $(2,2)$

So, given curve touch each other at $(2,2)$