The curves $x^3-3 x y^2=a$ and $3 x^2 y-y^3=b$, where a and b are constants, cut each other at an angle of
Answer & explanation
Correct answer: option 3
The equations of the two curves are
$C_1: x^3-3 x y^2=a$ ......(i)
$C_2: 3 x^2 y-y^3=b$ ......(ii)
Differentiating (i) and (ii) w.r.t. x, we get
$\left(\frac{d y}{d x}\right)_{C_1}=\frac{x^2-y^2}{2 x y}$ and $\left(\frac{d y}{d x}\right)_{C_2}=-\frac{2 x y}{x^2-y^2}$
Clearly, $\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}=-1$
So, the two curves intersect at right angle.