SOLUTION: Find the vertex, focus, length of latus rectum,and the equation of directrix of y^2 - 4y + 4x = 0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, focus, length of latus rectum,and the equation of directrix of y^2 - 4y + 4x = 0      Log On


   

Question 1119530: Find the vertex, focus, length of latus rectum,and the equation of directrix of y^2 - 4y + 4x = 0
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E2-4y%2B4x=0
-4x=y%5E2-4y
-4x%2B4=y%5E2-4y%2B4
highlight_green%28-4%28x-1%29=%28y-2%29%5E2%29-------horizontal axis of symmetry, parabola opens to the left; vertex is right-most point on the parabola.
-
-4=4p
p=1---------Directrix and focus both 1 unit away from the vertex. 
VERTEX       (1,2)
FOCUS         (0,2)
DIRECTRIX    x=2

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


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

x-h+=+%281%2F%284p%29%29%28y-k%29%5E2

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.

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|.

Put the given equation in that form:

y%5E2+-+4y+%2B+4x+=+0
y%5E2-4y%2B4+%2B+4x+-4+=+0
4x-4+=+-%28y%5E2-4y%2B4%29
4%28x-1%29+=+-1%28y-2%29%5E2
x-1+=+%281%2F-4%29%28y-2%29%5E2


This is in vertex form. The vertex is (1,2); p = -4/4 = -1.

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