Show that the curves `x=y^2` and `x y=k` cut at right angles, if `8k^2=1` .

Explanation

From `y = x^(2)`,
` (dy)/(dx)= 2x`
and from xy = k,
`x(dy)/(dx) + y = 0 rArr (dy)/(dx) = (-y)/x`
Let both curves intersect each other at right angles at point `(x_(1), y_(1))`.
`:. M_(1) = 2x_(1) and m_(2) = (-y_(1))/x_(1)`
` m_(1) m_(2) = - 1
` rArr 2x_(1) ((-y_(1))/(x_(1)))=-1 rArr y_(1) = 1/2`
From the curve y = `x^(2)`,
` y_(1)= x_(1)^(2)`
` rArr x_(1)^(2) = y_(1) = 1/2`
and from the curve xy = k,
` x_(1), y_(1) = k`
` rArr x_(1)^(2) y_(1)^(2) = k^(2)`
` rArr 1/2(1/2)^(2) = k^(2) rArr 8k^(2) = 1`