Discussion Forum

The two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ - 2 = 0 intersect at an angle of

Answer:

Explanation:

$x^{3}-3xy^{2} + 2 = 0$

differentiating w.r.t. x :

$3x^{2} -3x(2y)\frac{dy}{dx}-3y^{2} = 0$

$\Rightarrow\frac{dy}{dx}=\frac{3x^{2}-3y^{2}}{6xy}$ and $3x^{2}y-y^{3}-2 = 0$

differentiating w.r.t. x:

$3x^{2}\frac{dy}{dx}+6xy-3y^{2}\frac{dy}{dx}= 0$

$\Rightarrow\frac{dy}{dx} = -\left(\frac{6xy}{3x^{2}-3y^{2}}\right)$

Now, Product of slope

$\frac{3x^{2}-3y^{2}}{6xy}$× $-\left(\frac{6xy}{3x^{2}-3y^{2}}\right)$

.'. they are perpendicular. Hence, angle = π/2