Question 606915
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
Hi,
Yes, there is no y^2. This is a Parabola.
the vertex form of a parabola opening up or down, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex.
x^2+10x+4y+9=0
(x+5)^2 -25 + 4y + 9=0
(x+5)^2 + 4y -16 =0
{{{y = -(1/4)(x+5)^2 + 4}}}    a = -1/4<0, open downwards, Center(-5,4) Axis of symmetry x=-5 
The standard form is {{{(x +5)^2 = -4(y-4) = 4p(y -4)}}}, where  the focus is (h,k + p)
4p = -4, p = -1 , focus is (-5,3) and directrix is y = 5

{{{drawing(300,300,    -10,10,-10,10,   blue(line(-5,10,-5,-10))  
 grid(1),
circle(-5, 4,0.3),
circle(-5, 3,0.3),
graph( 300, 300,   -10,10,-10,10, 0,5,-(1/4)(x+5)^2 + 4  ))}}}

See below descriptions of various conics

Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius

 Standard Form of an Equation of an Ellipse is {{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1 }}} where Pt(h,k) is the center.
 a and b  are the respective vertices distances from center and ±{{{sqrt(a^2-b^2)}}}are the foci distances from center

Standard Form of an Equation of an Hyperbola opening right and  left is:
  {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} where Pt(h,k) is a center  with vertices 'a' units right and left of center.

Standard Form of an Equation of an Hyperbola opening up and down is:
  {{{(y-k)^2/b^2 - (x-h)^2/a^2 = 1}}} where Pt(h,k) is a center  with vertices 'b' units up and down from center.

the vertex form of a parabola opening up or down, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex.
The standard form is {{{(x -h)^2 = 4p(y -k)}}}, where  the focus is (h,k + p)

the vertex form of a parabola opening right or left, {{{x=a(y-k)^2 +h}}} where(h,k) is the vertex.
The standard form is {{{(y -k)^2 = 4p(x -h)}}}, where  the focus is (h +p,k )