Answer:case a) [tex]x^{2}=3y[/tex] ----> open upcase b) [tex]x^{2}=-10y[/tex] ----> open downcase c) [tex]y^{2}=-2x[/tex] ----> open leftcase d) [tex]y^{2}=6x[/tex] ----> open rightStep-by-step explanation:we know that1) The general equation of a vertical parabola is equal to[tex]y=a(x-h)^{2}+k[/tex]wherea is a coefficient(h,k) is the vertexIf a>0 ----> the parabola open upward and the vertex is a minimumIf a<0 ----> the parabola open downward and the vertex is a maximum2) The general equation of a horizontal parabola is equal to[tex]x=a(y-k)^{2}+h[/tex]wherea is a coefficient(h,k) is the vertexIf a>0 ----> the parabola open to the right If a<0 ----> the parabola open to the leftVerify each casecase a) we have[tex]x^{2}=3y[/tex]so[tex]y=(1/3)x^{2}[/tex][tex]a=(1/3)[/tex]so[tex]a>0[/tex]thereforeThe parabola open upcase b) we have [tex]x^{2}=-10y[/tex]so[tex]y=-(1/10)x^{2}[/tex][tex]a=-(1/10)[/tex][tex]a<0[/tex]thereforeThe parabola open downcase c) we have[tex]y^{2}=-2x[/tex]so[tex]x=-(1/2)y^{2}[/tex][tex]a=-(1/2)[/tex][tex]a<0[/tex]thereforeThe parabola open to the leftcase d) we have[tex]y^{2}=6x[/tex]so[tex]x=(1/6)y^{2}[/tex][tex]a=(1/6)[/tex][tex]a>0[/tex]thereforeThe parabola open to the right