Practicing Success
The two curves $x^3-3 x y^2+2=0$ and $3 x^2 y-y^3=2$ |
cut at right angles touch each other cut at an angle $\pi / 3$ cut at an angle $\pi / 4$ |
cut at right angles |
We have, $x^3-3 x y^2+2=0$ .......(i) and, $3 x^2 y-y^3-2=0$ ......(ii) Differentiating (i) and (ii) with respect to x, we obtain $\left(\frac{d y}{d x}\right)_{C_1}=\frac{x^2-y^2}{2 x y}$ and $\left(\frac{d y}{d x}\right)_{C_2}=\frac{-2 x y}{x^2-y^2}$ Clearly, $\left(\frac{d y}{d x}\right)_{C_1} \times\left(\frac{d y}{d x}\right)_{C_2}=-1$ Hence, the two curves cut at right angles. |