Question

The two curves $x^3-3x{y}^2+5=0$ and $3x^{2}y-y^3-7=0$

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

Answer 1

Answer: (A)

Differentiating $x^3-3x{y}^2+5=0$, we get
$3x^2-3y^2-6xy\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{x^2-y^2}{2xy}$
Differentiating $3x^{2}y-y^3-7-0$, we get
$6xy+3x^2\frac{dy}{dx}-3y^2\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{2xy}{y^2-x^2}$
Since, product of slopes is $\frac{x^2-y^2}{2xy}\times \frac{2xy}{y^2-x^2}=-1$
$\therefore$ The two curves cut at right angle.