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what is the length of the focal width x=1/4y^2
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@jim_thompson5910
x = (1/4)*y^2 4x = y^2 4(x-0) = (y-0)^2 4*1*(x-0) = (y-0)^2
the equation 4*1*(x-0) = (y-0)^2 is in the form 4p(x-h) = (y-k)^2 where p is the focal distance (h,k) is the vertex so we can see that p = 1
p = 1 means that the distance from the focus to the vertex is 1 unit
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is the focal point=1?
it's not at that location, but that is the focal distance
what is the focal width?
oh I'm mixing up focal distance and focal width
luckily the two are found using p
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|p| = focal distance |4p| = focal width
4 is the answer?
yes
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