Practicing Success
The two curves $x^3-3 x y^2+5=0$ and $3 x^2 y-y^3-7=0$ |
cut at right angles touch each other cut at an angle $\frac{\pi}{4}$ cut at an angle $\frac{\pi}{3}$ |
cut at right angles |
Differentiating $x^3-3 x y^2+5=0$, we get $3 x^2-3 y^2-6 x y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x^2-y^2}{2 x y}$ Differentiating $3 x^2 y-y^3-7=0$, we get $6 x y+3 x^2 \frac{d y}{d x}-3 y^2 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x y}{y^2-x^2}$ Since, product of slopes is $\frac{x^2-y^2}{2 x y} \times \frac{2 x y}{y^2-x^2}=-1$ ∴ The two curves cut at right angle. |