What is the angle between these two curves `x^3-3xy^2+2=0` and `3x^2y-y^3-2=0`

Options

A. cut at right angles
B. touch each other
C. cut at an angle ` pi//3 `
D. cut at an angle `pi//4`

Answer

A

Explanation

We have,
` x^(3)-3xy^(2)+2 =0 " "...(i) `
and, ` 3x^(2)y-y^(3)-2=0 " "...(ii)`
Differentiating (i) and (ii) with respect to x, we obtain
` ((dy)/(dx))_(C_(1)) =(x^(2)-y^(2))/(2xy) " and " ((dy)/(dx))_(C_(2))=(-2xy)/(x^(2)-y^(2)) `
Clearly, ` ((dy)/(dx))_(C_(1))xx((dy)/(dx))_(C_(2))=-1`
Hence, the two curves cut at right angles.