Answer
$x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2$
Work Step by Step
$x^2 +(y+\dfrac{g \ t^2}{2})=(v_0 \cos \alpha)^2 t^2+(v_0 \sin \alpha)^2 t^2$
$\implies x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2 ( \cos^2 \alpha+ \sin^2 \alpha)$
$\implies x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2 (1)$
$\implies x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2$
