Question

In \( \mathbb{R}^2 \), the family of trajectories orthogonal to the family of asteroids \( x^{2/3} + y^{2/3} = a^{2/3} \) is given by

• A. \( x^{4/3} + y^{4/3} = c^{4/3} \)
• B. \( x^{4/3} - y^{4/3} = c^{4/3} \)
• C. \( x^{5/3} - y^{5/3} = c^{5/3} \)
• D. \( x^{2/3} - y^{2/3} = c^{2/3} \)

Hint:

To find the orthogonal trajectories of a given family of curves, differentiate implicitly and use the orthogonality condition that the slopes of the curves must multiply to \( -1 \).

Answer 1

The Correct Option is B

Step 1: Find the differential equation of the given family

Differentiating both sides with respect to $x$: $$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0$$

$$x^{-1/3} + y^{-1/3}\frac{dy}{dx} = 0$$

$$y^{-1/3}\frac{dy}{dx} = -x^{-1/3}$$

$$\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{1/3}$$

Step 2: Find the differential equation of orthogonal trajectories

For orthogonal trajectories, we replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (or equivalently, $\frac{dy}{dx}$ with $-\frac{1}{dy/dx}$):

$$\frac{dy}{dx} = -\frac{1}{-\left(\frac{y}{x}\right)^{1/3}} = \left(\frac{y}{x}\right)^{-1/3} = \left(\frac{x}{y}\right)^{1/3}$$

Step 3: Solve the differential equation

$$\frac{dy}{dx} = \left(\frac{x}{y}\right)^{1/3}$$

$$y^{1/3}dy = x^{1/3}dx$$

Integrating both sides: $$\int y^{1/3}dy = \int x^{1/3}dx$$

$$\frac{y^{4/3}}{4/3} = \frac{x^{4/3}}{4/3} + C_1$$

$$\frac{3}{4}y^{4/3} = \frac{3}{4}x^{4/3} + C_1$$

$$y^{4/3} = x^{4/3} + C$$

where $C = \frac{4C_1}{3}$

This can be rewritten as: $$x^{4/3} - y^{4/3} = -C$$

Or equivalently (letting $c^{4/3} = -C$): $$x^{4/3} - y^{4/3} = c^{4/3}$$

Answer: (B)