For slope of curve x3 + kxy2 = -2, we differentiate
its equation w.r.t. x then
3x2 + ky2 + 2kxy\(\frac{dy}{dx}=0\)
⇒ \(\frac{dy}{dx}=-\frac{(3x^2+ky^2)}{2kxy}\) = m1 (Let)--(1)
For slope of curve 3x2y - y3 = 2, we differentiate its equation w.r.t. x then
(3x2 - 3y2)\(\frac{dy}{dx}\) + 6xy = 0
⇒ \(\frac{dy}{dx}\) = \(\frac{-6xy}{3x^2-3y^2}\) = m2(Let)--(2)
Since, both curves are orthogonal to each other
∴ m1m2 = -1
⇒ \(\frac{-(3x^2+ky^2)}{2kxy}\times\frac{-6xy}{3x^2-3y^2}=-1\)
⇒ \(\frac{3(3x^2+ky^2)}{k(3x^2-3y^2)}=-1\)
⇒ 3x2 + ky2 = -kx2 + ky2
⇒ -kx2 = 3x2
⇒ k = -3
∴ |k| = 3