Question 1119530
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The equation has a y^2 term, so the parabola opens right or left.  Vertex form for the equation of a parabola that opens right or left is<br>
{{{x-h = (1/(4p))(y-k)^2}}}<br>
where the vertex is (h,k) and p is the directed distance from the directrix to the vertex and from the vertex to the focus.<br>
Note with this form of the equation, the length of the latus rectum (perpendicular to the axis of symmetry and through the focus) is |4p|.<br>
Put the given equation in that form:<br>
{{{y^2 - 4y + 4x = 0}}}
{{{y^2-4y+4 + 4x -4 = 0}}}
{{{4x-4 = -(y^2-4y+4)}}}
{{{4(x-1) = -1(y-2)^2}}}
{{{x-1 = (1/-4)(y-2)^2}}}
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This is in vertex form.  The vertex is (1,2); p = -4/4 = -1.<br>
vertex: (1,2)
focus: p = 1 left of the vertex; at (0,2)
directrix: p = 1 right of the vertex; x = 2
length of latus rectum: |4p| = 4