Question

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

Answer 1

Given curves are

$xy=4$ ......$\left(1\right)$

$x^2+y^2=8$ .....$\left(2\right)$

We need to find the point of intersection of curves.

From $\left(1\right)$ $y=\frac{4}{x}$

Substituting this in $\left(2\right)$ we get,

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

$\Rightarrow x^4-8x^2+16=0$

$\Rightarrow (x^2-4{)}^2=0$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

Using $\left(1\right)$ we get $y=\pm 2$

Therefore points of intersection are $(2,2),(-2,-2)$

Differentiating curve 1 we get

$\Rightarrow y+x\left(\frac{dy}{dx}\right)=0$

$\Rightarrow \left(\frac{dy}{dx}\right)=\frac{-y}{x}$

At $(2,2),(-2,-2)$ $\Rightarrow \left(\frac{dy}{dx}\right)=-1$

Differentiating curve 2 we get

$\Rightarrow 2x+2y\left(\frac{dy}{dx}\right)=0$

$\Rightarrow \left(\frac{dy}{dx}\right)=\frac{-x}{y}$

At $(2,2),(-2,-2)$ $\Rightarrow \left(\frac{dy}{dx}\right)=-1$

Thus we can say that the 2 curves touch each other at their point of intersection.