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If curves \( x^{3}+k x y^{2}=-2 \) and \( 3 x^{2} y-y^{3}=2 \) are orthogonal to each other then \( |k| \) is

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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

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