Question 79487
Identify the conic section represented by each equation. If it is a parabola, give the vertex. If it is a circle, give the center and radius. If it is an ellipse or a hyberbola, give the center and foci. Sketch the graph.

{{{3y^2-x-6y+5=0}}}

It is a parabola because it only has ONE squared letter.

Get all the terms in the squared letter on the left and 
all other terms on the right:

{{{3y^2-6y=x-5}}}

Factor out the coefficient of {{{y^2}}} on the left:

{{{3(y^2-2y) = x-5}}}

Complete the square by adding 1 inside the parentheses
on the left.  However note that when we placing a 1 
inside the parentheses on the left, we are really adding 3
because of the 3 multiplier in front of the parentheses
so we must add 3 to the right side to offset.

{{{3(y^2-2y+1) = x-5+3}}}

Factor the trinomial on the left side as a perfect square,
and combine terms on the right:

{{{3(y-1)^2 = x-2}}}

Multiply both sides by {{{1/3}}}

{{{(1/3)3(y-1)^2 = (1/3)(x-2)}}}

{{{1(y-1)^2 = (1/3)(x-2)}}}

{{{(y-1)^2 = (1/3)(x-2)}}}

Now the equation is in the form:

{{{(y-k)^2 = 4p(x-h)}}}

where the vertex is (h,k) = (2,1)

{{{graph(300,167, -2,7,-2,3,0,0,1+sqrt((x-2)/3),0,0,0,1-sqrt((x-2)/3))}}}