The two curves `x^3-3xy^2+2=0` and `3x^2y-y^3-2=0`

Options

A. `pi/4`
B. `pi/3`
C. `pi/2`
D. `pi/6`

Answer

C

Explanation

Equation of two curves are given by
`x^(3)-3xy^(2)+2=0`
and `3x^(2)y-y^(3)-2=0` [on differentiating w.r.t. x]
`rArr 3x^(2)-3[x.2y(dy)/(dx)+y^(2).1]+0=0`
and `3y^(2)(dy)/(dx) = 3x^(2)(dy)/(dx)+6xy`
`rArr (dy)/(dx)=(3x^(2)-3y^(2))/(6xy)`
and `(dy)/(dx)=(6xy)/(3y^(2)-3x^(2))`
`rArr (dy)/(dx)= (3(x^(2))-y^(2))/(6xy)`
and `(dy)/(dx) = (-6xy)/(3(x^(2)-y^(2))`
`rArr m_(1)=(x^(2)-y^(2))/(2xy)`
and `m_(1)=(x^(2)-y^(2))/(2xy)`
and `m_(2)=(-2xy)/(x^(2)-y^(2))`
`therefore m_(1)m_(2)=(x^(2)-y^(2))/(2xy) -(2xy)/(x^(2)-y^(2))=-1`
Hence, both the curves are intersecting at right angles i.e., making `pi/2` with each other.