# In which region the curve $y^{2}(a+x)=x^{2}(3a-x)$ does not lie?

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

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