The number of tangents to the curve $y^2(x-a)=x^2(x+a)(a>0)$ that are parallel to the X-axis is ……

Equation of given curve is $y^2(x-a)=x^2(x+a)\Rightarrow y^2(x-a)=x^3+a{x}^2\to \left(1\right)$

Let $P=(x,y)$ be a point on the curve.

$\left(1\right)\Rightarrow y^{2}1+(x-a)2y\frac{dy}{dx}=3x^2+2ax\Rightarrow 2(x-a)y\frac{dy}{dx}=3x^2+2ax-y^2$

$\Rightarrow \frac{dy}{dx}=\frac{3x^2-y^2+2ax}{2(x-a)y}\Rightarrow m={\left(\frac{{\displaystyle dy}}{{\displaystyle dx}}\right)}_p=\frac{3x^2-y^2+2ax}{2(x-a)y}$

The tangent at ‘P’ is parallel to x-axis $\Rightarrow m=0\Rightarrow 3x^2-y^2+2ax=0\Rightarrow y^2=3x^2+2ax-\left(2\right)$

From(1) &(2), $(3x^2+2ax)(x-a)=x^3+a{x}^2\Rightarrow 3x^3+2a{x}^2-3a{x}^2-2a^{2}x=x^3+a{x}^2$

$\Rightarrow 2x^3-2a{x}^2-2a^{2}x=0\Rightarrow x=0,x^2-ax-a^2=0$

If $x=0$ then from equation (1) , $y^2(0-a)=0\Rightarrow y=0.\therefore P=(0,0)$

Moreover $x^2-ax-a^2=0$ has imaginary roots.

Hence there exists one tangent to the curve at ‘P’.