Find the equation of the parabola that has vertex,(-2,4),axis of parabola is parallel to the x-axis and passing through the point (3,8)?

##16x+40y-5y^2=48##

If the axis of symmetry of a parabola is parallel to the X-axis then the vertex form of the equation is ##color(white)("XXX")x=m(y-b)^2+a## with the vertex at ##(a,b)## for some constant ##m##

Given the vertex at ##(-2,4)## this becomes ##color(white)("XXX")x=m(y-4)^2+(-2)##

Since this equation has a solution ##(x,y)=(3,8)## we have ##color(white)("XXX")3=m(8-4)^2-2 ##color(white)("XXX")16m=5 color(white)("XXX")m=5/16

So the equation is ##color(white)("XXX")x=5/16(y-4)^2-2##

Which could be simplified in various forms: ##color(white)("XXX")16x=5(y^2-8y+16)-32##

##color(white)("XXX")16x=5y^2-40y+48##

##color(white)("XXX")16x+40y-5y^2=48## graph{16x+40y-5y^2-48=0 [-4.07, 8.414, 2, 8.5]}