Question

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

• 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 $x^3 - 3xy^2 + 5 = 0$, 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.