\[\begin{array}{1 1}(1)\;x>0&(2)\;0< x < 3a \\(3)\;x<-a\; \text{and}\; x>3a & (4)-a \end{array} \]

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$y^2 =\large\frac{x^2 (3a-x)}{a+x}$

When $ x < -a$

$x+a < 0$

and $3a-x > 0$

$y^2$ is -ve

y is imaginary

$\therefore$ the curve does not exist in this region.

When $ x > 3a $

$3a-x > 0$

and $a+x > 0$

$y^2$ is -ve

y is imaginary

$\therefore$ the curve does not exist in this region.

When $ -a < x <3a$

Say $x=0$

$x+a=0+a >0 $ and $3a-x=3a-0=3a > 0$

$y^3$ is positive

y is real

The curve exist in these region

Hence 3 is the correct answer.

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