Question

Two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2
(a) touch each other (b) cut at right angle
(c) cut an angle $\frac{\mathrm{\pi }}{3}$ (d) cut at an angle $\frac{\mathrm{\pi }}{4}$

Answer 1

The given curves are x³ − 3xy² + 2 = 0 and 3x²y − y³ = 2.

We know that the angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.

Let (x₁, y₁) be the point of intersection of two curves,

Consider the first curve C₁ ≡ x³ − 3xy² + 2 = 0.

x³ − 3xy² + 2 = 0

Differentiating both sides with respect to x, we get

$3x^2-3\left(x\times 2y\frac{dy}{dx}+y^2\times 1\right)+0=0$

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

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

∴ Slope of tangent to the first curve at (x₁, y₁) = ${\left(\frac{dy}{dx}\right)}_{C_1}=\frac{x_1^2-y_1^2}{2x_1{y}_1}$

Now, consider the second curve C₂ ≡ 3x²y − y³ = 2.

3x²y − y³ = 2

Differentiating both sides with respect to x, we get

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

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

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

∴ Slope of tangent to the second curve at (x₁, y₁) = ${\left(\frac{dy}{dx}\right)}_{C_2}=-\frac{2x_1{y}_1}{x_1^2-y_1^2}$

Now, ${\left(\frac{dy}{dx}\right)}_{C_1}\times {\left(\frac{dy}{dx}\right)}_{C_2}=\left(\frac{x_1^2-y_1^2}{2x_1{y}_1}\right)\times \left(-\frac{2x_1{y}_1}{x_1^2-y_1^2}\right)=-1$

The tangents to the two curves are perpendicular to each other.

Thus, the two curves cut at right angle.

Hence, the correct answer is option (b).