Q. 2 B5.0( 1 Vote )

# Show that the fol

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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