SOLUTION: Could someone please help me with writing a conic equation in standard form? The question is: A parabola with vertex at (5,0) and directrix at x=7. I have (y-k)^2=-4p(x-h) (y-0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Could someone please help me with writing a conic equation in standard form? The question is: A parabola with vertex at (5,0) and directrix at x=7. I have (y-k)^2=-4p(x-h) (y-0      Log On


   

Question 1028741: Could someone please help me with writing a conic equation in standard form? The question is: A parabola with vertex at (5,0) and directrix at x=7.
I have (y-k)^2=-4p(x-h)
(y-0)^2=-4p(x-5)
p=2
y^2=-8p(x-5)--not sure if this is correct? Thanks!
Found 2 solutions by josgarithmetic, Natolino1983:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The FOCUS is on the other side of the vertex from the directrix. Directrix is 2 units away from the vertex, so focus is therefore (3,0).

The basic form you chose is the right kind, which you would get or be able, if you were deriving the equation for your parabola using the definition of Parabola and Distance Formula.

%28y-0%29%5E2=-4p%28x-5%29, and you are supposed to understand -4p=-4%2A2, meaning p=2. Combining those, y%5E2=-8%28x-5%29. Continuing toward standard form,
%281%2F8%29y%5E2=-8x%2B40
-%281%2F8%29y%5E2=x-5
highlight%28x=-%281%2F8%29y%5E2%2B5%29

Answer by Natolino1983(23) About Me  (Show Source):
You can put this solution on YOUR website!
You almost got the answer. In this case, p = 2 (distance between vertex and directrix).
So the equation is y^2 = -8×(x - 5), because -4×p = -4×2 =-8.