9. When $\theta$ varies, the equation of the locus of the vertex of the parabola $y=x^{2}+2 x \sin \theta+1$ is
9. $y=-x^{2}+1, -1 \leqslant x \leqslant 1$. Hint: The original equation is $y=x^{2}+2 x \sin \theta+1=(x+\sin \theta)^{2}+\cos ^{2} \theta$, let the vertex coordinates be $(x, y)$, then $x=-\sin \theta, y=\cos ^{2} \theta$, eliminate $\theta$, i.e., $y=-x^{2}+1$.