# Prove that the curves $xy=4$ and $x^2+y^2=8$ touch each other.

Toolbox:
• If $y=f(x),$ then $\bigg(\large\frac{dy}{dx}\bigg)_p$ = slope of the tangent to $y=f(x)$ at point $p$
Step 1
The given curves are
$xy=4\: \: \: (1) \Rightarrow y=\large\frac{4}{x}$
$x^2+y^2=8\: \: \: (2)$
Substitute the value of $y$ in equation (2)
$x^2+ \bigg( \large\frac{4}{x} \bigg)^2=8$
$\Rightarrow \large\frac{x^4+16}{x^2}=8$
$\Rightarrow x^4+16=8x^2$
$\Rightarrow x^4-8x^2+16=0$
$\Rightarrow (x^2-4)^2=0$
$\Rightarrow x= \pm 2$
When $x=2,\: y=2\: and \: x=-2,\: y=-2$
Hence the two curves intersect at $p(2,2)\: and \: q(-2, -2)$
Step 2
Now $xy=4$
On differentiating w.r.t $x$ we get,
$x\large\frac{dy}{dx}+y=0 \Rightarrow \large\frac{dy}{dx}=-\large\frac{y}{x}$
Consider $x^2+y^2=8$
On differentiating w.r.t $x$ we get,
$2x+2y\large\frac{dy}{dx}=0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{x}{y}$
At $p(2,2)$ we have
For curve (1) $\bigg( \large\frac{dy}{dx} \bigg)_{(c1)}=-\large\frac{2}{2}=-1$
For curve (2) $\bigg( \large\frac{dy}{dx} \bigg)_{(c2)}=-\large\frac{2}{2}=-1$
Hence the $\bigg(\large\frac{dy}{dx} \bigg)_{c1}= \bigg(\large\frac{dy}{dx}\bigg)_{c2}$ at $p$
So the two curves touch eachother.