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Answer :

Given:


Curves x3 – 3xy2 = – 2 ...(1)


& 3x2y – y3 = 2 ...(2)


Adding (1) & (2),we get


x3 – 3xy2 + 3x2y – y3 = – 2 + 2


x3 – 3xy2 + 3x2y – y3 = – 0


(x – y)3 = 0


(x – y) = 0


x = y


Substituting x = y on x3 – 3xy2 = – 2


x3 – 3×x×x2 = – 2


x3 – 3x3 = – 2


– 2x3 = – 2


x3 = 1


x = 1


Since x = y


y = 1


The point of intersection of two curves is (1,1)


First curve x3 – 3xy2 = – 2


Differentiating above w.r.t x,


3x2 – 3(1×y2 + x×2y) = 0


3x2 – 3y2 – 6xy0


3x2 – 3y2 = 6xy




m1 ...(3)


Second curve 3x2y – y3 = 2


Differentiating above w.r.t x,


3(2x×y + x2×) – 3y20


6xy + 3x23y20


6xy + (3x2 – 3y2)0




m2 ...(4)


When m1 & m2 =



× = – 1


Two curves x3 – 3xy2 = – 2 & 3x2y – y3 = 2 intersect orthogonally.


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