${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha$ $\Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha$ $\Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha$ $\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}}$ squaring both sides ${x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha$ $\therefore$ 4x 2 – 4xy cos $\alpha$ + y 2 = 4 ${\sin ^2}\alpha$