Given,

${\text{cos}}^{-1}\left(\text{x}\right)-{\text{cos}}^{-1}\left(\frac{\text{y}}{2}\right)=\alpha ...\left(i\right)$         
let  ${\text{A=cos}}^{-1}\left(\text{x}\right)\text{, }\text{B= }{\text{cos}}^{-1}\left(\frac{\text{y}}{2}\right)$
$\Rightarrow$   $\cos A=x, \cos B=\frac{\text{y}}{2} ...\left(ii\right)$                
Using equation $\left(i\right)$ $\Rightarrow \text{A}-\text{B}=\alpha$
       $\Rightarrow \text{cos}\left(\text{A}-\text{B}\right)=\text{cos }\alpha$
       $\Rightarrow \text{cos}\text{A}\text{ cos B}+\text{sin A}\text{ sin B}=\text{cos }\alpha$
       $\Rightarrow \frac{\text{xy}}{2}+\sqrt{1-{\text{x}}^2} \sqrt{1-\frac{{\text{y}}^2}{4}}=\text{cos }\alpha$   Using equation $\left(ii\right)$
$\Rightarrow {\left(\text{cos}\alpha -\frac{\text{xy}}{2}\right)}^2=\left(1-{\text{x}}^2\right) \left(1-\frac{{\text{y}}^2}{4}\right)$
$\Rightarrow {\text{cos}}^2\alpha +\frac{{\text{x}}^2{\text{y}}^2}{4}-\text{xy}\text{ cos}\mathrm{ \alpha }=1-\frac{{\text{y}}^2}{4}-{\text{x}}^2+\frac{{\text{x}}^2{\text{y}}^2}{4}$

$\Rightarrow \frac{{\text{y}}^2}{4}+{\text{x}}^2-\text{xy cos} \alpha =1-{\text{cos}}^2\alpha$
$\Rightarrow {\text{x}}^2+\frac{{\text{y}}^2}{4}-\text{xy }\text{cos }\alpha ={\text{sin}}^2\alpha$
$\Rightarrow 4{\text{x}}^2-4\text{xy }\text{cos }\alpha +{\text{y}}^2=4{\text{sin}}^2\alpha$ .