If the equation given is #y=2x^2# and the point given is (-1,2), how would you find the p value from the form #(x-0)^2=4p(y-k)#?

1 Answer
Jake M.
Aug 8, 2018

#p=1/8#

Explanation:

We can rearrange the formula:
#x^2 = 4p(y-k) implies y = x^2/(4p) + k#

Therefore, we know that the coefficient of #x^2# is #2#, hence
#2 = 1/(4p) implies p = 1/8 #

We could also have just done this with the given point by plugging in the known values of #x,y# that work:
#x^2 = 4p(y-k) #
#(-1)^2 = 4p(2-k) implies 1 = (8- 4k)p#
Since we can see that the vertical shift is 0 (since the function also intercepts 0), we get #k=0# i.e. #1 = 8p implies p = 1/8#.

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