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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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Answer» 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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