If t in R - {-1},then the locus of the point | vedclass

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(A) Given $x = \frac{3at}{1+t^3}$ and $y = \frac{3at^2}{1+t^3}$.We observe that $xy = \frac{9a^2t^3}{(1+t^3)^2}$ and $x^2y = \frac{27a^3t^4}{(1+t^3)^3

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$x^3+y^3=3axy$

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(A) Given $x = \frac{3at}{1+t^3}$ and $y = \frac{3at^2}{1+t^3}$.We observe that $xy = \frac{9a^2t^3}{(1+t^3)^2}$ and $x^2y = \frac{27a^3t^4}{(1+t^3)^3}$,which is not the direct path.Instead,consider $x^3 + y^3 = \left(\frac{3at}{1+t^3}\right)^3 + \left(\frac{3at^2}{1+t^3}\right)^3$.$x^3 + y^3 = \frac{27a^3t^3 + 27a^3t^6}{(1+t^3)^3} = \frac{27a^3t^3(1+t^3)}{(1+t^3)^3} = \frac{27a^3t^3}{(1+t^3)^2}$.Now,$3axy = 3a \left(\frac{3at}{1+t^3}\right) \left(\frac{3at^2}{1+t^3}\right) = \frac{27a^3t^3}{(1+t^3)^2}$.Therefore,$x^3 + y^3 = 3axy$.

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

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