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Question 982005: Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.
a. y^2 = -8x
b. 16y = x^2
c. y = x^2
d. x = y^2
I tried to solve this problem and got y^2= -16 but thats not an answer choice. Please help! Is it y^2 = -8x?
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! Find the standard form of the equation of the parabola with a focus at (-4, 0) and a directrix at x = 4.
I won't do that exact problem but one exactly like it so you can use it as a
model to do yours by. I'll do this one instead:
Find the standard form of the equation of the parabola with a focus at (-3, 0) and a directrix at x = 3.
See the graph below. I plot the focus, draw the directrix (in green). And
since I know that thevertex is halfway between the focus and the directrix, I
know that the vertex is (0,0). I know that the focal distance (from the vertex
to the focus is -3 (negative because you have to go left from the vertex to the
focus. I also know that a parabola that opens left or right has the equation
(y-k)² = 4p(x-h)
where (h,k) is the vertex and p = the focal distance. So I
substitute h=0, k=0, p=-3, and get
(y-0)² = 4(-3)(x-0)
y² = -12x
Now do yours the exact same way.
Edwin
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website!
Hello, I think the correct answer is not listed. Sometimes books and computer
modules have errors. You are almost right with your y^2 = -16 but you left off
something. What was it?
Phil
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