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If $t \in \mathbb{R}-\{-1\}$, then the locus of the point $\left(\frac{3 a t}{1+t^3}, \frac{3 \mathrm{at}^2}{1+t^3}\right)$ is
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The correct answer is:
$x^3+y^3=3 a x y$
Here, $x=\frac{3 a t}{1+t^3}, y=\frac{3 a t^2}{1+t^3}$
$\begin{aligned} & x^3+y^3=\left(\frac{3 a t}{1+t^3}\right)^3+\left(\frac{3 a t^2}{1+t^3}\right)^3 \\ & =\left(\frac{3 a t}{1+t^3}\right)^3\left(1+\mathrm{t}^3\right) \\ & =\frac{(3 a t)^3}{\left(1+\mathrm{t}^3\right)^2}=\frac{3 a \cdot 3 a t \cdot 3 a t^2}{\left(1+\mathrm{t}^3\right)^2}=3 a \cdot x \cdot y\end{aligned}$
$\begin{aligned} & x^3+y^3=\left(\frac{3 a t}{1+t^3}\right)^3+\left(\frac{3 a t^2}{1+t^3}\right)^3 \\ & =\left(\frac{3 a t}{1+t^3}\right)^3\left(1+\mathrm{t}^3\right) \\ & =\frac{(3 a t)^3}{\left(1+\mathrm{t}^3\right)^2}=\frac{3 a \cdot 3 a t \cdot 3 a t^2}{\left(1+\mathrm{t}^3\right)^2}=3 a \cdot x \cdot y\end{aligned}$
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