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

Explanation

Given, eqution of curves are
and `xy=4`…………..(i)
`rArr x.(dy)/(dx) + y=0` ……………..(ii)
and `x.(dy)/(dx) + y=0`
`rArr 2x+2y(dy)/(dx)=0`
`rArr (dy)/(dx)=-y/x`
and `(dy)/(dx) = -y/x`
`(dy)/(dx) = (-2x)/(2y)`
and `(dy)/(dx) = -y/x=m_(1)` [say]
and `(dy)/(dx) = -x/y = m_(2)` [say]
Since, both the curves should have same slope.
`-y/x = -x/y rArr -y^(2)=-x^(2)`
`rArr x^(2)=y^(2)`...............(iii) Using the value of `x^(2)` in Eq. (ii), we get
`y^(2) + y^(2)=8`
`rArr y^(2)=4 rArr y= +-2`
For y=2, `x=4/2=2`
and for y=`-2`, x`=4/-2 = -2`
Thus, the required points of intersection are (2,2) and `(-2,-2)`.
For (2,2), `m_(1) = -y/x = -2/2=-1`
and `m_(2) = -x/y=-2/2=-1`
`therefore m_(1) = m_(2)`
For `(-2,-2)` `m_(1)=-y/x = -(-2)/(-2)=-1`
and `m_(2) = -x/y=-(-2)/(-2)=-1`
Thus, for both the intersection points, we see that slope of both the curves are same.
Hence, the curves touch each other.