Prove that the curve `y = x^2` and `xy = k` intersect orthogonally if `8k^2 = 1`.

Explanation

Given, equation of curves are `=2x=y^(2)` ………….(i)
and `2xy=k`……………….(ii)
`rArr y=k/(2x)` [From eq.(ii)]
From Eq.(i), `2x=(k/(2x))^(2)`
`rArr 8x^(3)=k^(2)`
`rArr x^(3)=1/8k^(2)`
`rArr x=1/2k^(2//3)`
`rArr y=k/(2x) = k/(2.1/2k^(2//3)) = k^(1//3)`
Thus, we get point of intersection of curves which is `(1/2k^(2//3), k^(1//3))`
From Eqs. (i) and (ii), `2=2y(dy)/(dx)`
and `2[x.(dy)/(dx) + y.1]=0`
`rArr (dy)/(dx) =1/y`
and `(dy)/(dx) =(-2y)/(2x) =-y/3`
`rArr (dy)/(dx) = (-2y)/(2x)=-y/x`
`rArr (dy)/(dx)_(1/2k^(2//3),k^(1//3)) = 1/k^(1//3)` [say `m_(1)`]
and `(dx)/(dy)_(1/2k^(2//3),k^(1//3)) = k^(1//3)/(1/2k^(2//3)) = -2k^(-1//3)` [say `m_(2)`]
Since, the curves intersect orthogonally.
i.e., `m_(1).m_(2)=-1`
`rArr 1/k^(1//3).(-2k^(-1//3))=-1`
`rArr -2k^(-2//3) = -1`
`rArr 2/(k^(2//3)) =1`
`k^(2//3)=2`
`therefore k^(2)=8`
Which is the required condition.