Keep the trigonometric range and solve an intersection exactly

$\sin^2\theta+\cos^2\theta=1$ gives $y/2+x^2=1$, so $y=2(1-x^2)$ with $-1\le x\le1$. The unrestricted parabola would include points impossible for $x=\cos\theta$.

$x=0$ gives $t(t^2+1)=0$, hence $(0,-1)$. For $y=0$, $t=1/2$ and $x=1/8+1/2=5/8=0.625$, hence $(5/8,0)$. The closed endpoints are $(-2,-3)$ at $t=-1$ and $(2,1)$ at $t=1$. Since $dx/dt=3t^2+1>0$ and $dy/dt=2$, the curve increases throughout. Its slope is largest at $t=0$, giving the smooth bend through $(0,-1)$.

At $y=1/2$, $1/2=2t-1$ gives $t=3/4$, which is allowed. Then $x=(3/4)^3+3/4=27/64+48/64=75/64$. The intersection is $(75/64,1/2)$.