Answer
$y=x\tan \alpha -16\dfrac {x^{2}}{v^{2}_{0}\cos ^{2}\alpha }=x\tan \alpha -16\dfrac {x^{2}}{v^{2}_{0}}\left( 1+\tan ^{2}\alpha \right) $
Work Step by Step
$x=\left( v_{0}\cos \alpha \right) t\Rightarrow t=\dfrac {x}{v_{0}\cos \alpha }$
Lets eliminate t in second equation
$y=\left( v_{0}\sin \alpha \right) {t}-16t^{2}=\left( v_{0}\sin \alpha \right) \dfrac {x}{v_{0}\cos \alpha }-16\left( \dfrac {x}{v_{0}\cos \alpha }\right) ^{2}=x\tan \alpha -16\dfrac {x^{2}}{v^{2}_{0}\cos ^{2}\alpha }=x\tan \alpha -16\dfrac {x^{2}}{v^{2}_{0}}\left( 1+\tan ^{2}\alpha \right) $
