Question

If \( t \in \mathbb{R} \setminus \{-1\} \), then the locus of the point \[ \left( \frac{3at}{1 + t^3}, \frac{3at^2}{1 + t^3} \right) \] is:

• A. \( x^3 + y^3 = 3a x^2 y^2 \)
• B. \( x^3 - 3ax^2 y - 3axy^2 + y^3 = 0 \)
• C. \( x^3 + y^3 = 3axy \)
• D. \( x^3 - y^3 = 3axy \)

Hint:

Use parametric equations to eliminate the parameter and derive the Cartesian form.

Answer 1

The Correct Option is C

Given: \[ x = \frac{3at}{1 + t^3}, \quad y = \frac{3at^2}{1 + t^3} \] Let’s clear denominators and consider: \[ x = \frac{3at}{1 + t^3}, \quad y = \frac{3at^2}{1 + t^3} \Rightarrow x^3 + y^3 = \frac{27a^3(t^3 + t^6)}{(1 + t^3)^3} \] Now: \[ 3axy = 3a \cdot \frac{3at}{1 + t^3} \cdot \frac{3at^2}{1 + t^3} = \frac{27a^3t^3}{(1 + t^3)^2} \] Hence: \[ x^3 + y^3 = 3axy \]