SOLUTION: Find the standard form of the equation of the parabola with vertex at (5,2) and focus at (3,2).

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Question 1087919: Find the standard form of the equation of the parabola with vertex at (5,2) and focus at (3,2).
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
(5,2) distance from (3,2) is 2 units. Directrix x=7.
Symmetry axis, y=2. Concave to the left.

-4%2A2%2Ax=%28y-5%29%5E2%2B2
highlight_green%28-8x=%28y-6%29%5E2%2B2%29
-
-8x=y%5E2-12y%2B36%2B2
-8x=y%5E2-12y%2B38
highlight_green%28x=-%281%2F8%29y%5E2%2B%283%2F2%29y-19%2F4%29

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!
Find the standard form of the equation of the parabola with vertex at (5,2) and focus at (3,2).
Since the focus: (3, 2) is to the LEFT of the VERTEX (5, 2), this is a PARABOLA with a HORIZONTAL AXIS of SYMMETRY. 
Therefore, the CONIC form of the PARABOLA with a HORIZONTAL AXIS, or %28y+-+k%29%5E2+=+4p%28x+-+h%29 is used.
h = 5; k = 2 p = – 2
%28y+-+k%29%5E2+=+4p%28x+-+h%29 becomes:
%28y+-+2%29%5E2+=+4%28-+2%29%28x+-+5%29
y%5E2+-+4y+%2B+4+=+-+8x+%2B+40
8x+=+-+y%5E2+%2B+4y+%2B+36
x+=+%28-+1%2F8%29y%5E2+%2B+%284%2F8%29y+%2B+36%2F8
<======== Equation of parabola