SOLUTION: Find the focal diameter for the parabola 34.8y = x^2

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Question 1204030: Find the focal diameter for the parabola 34.8y = x^2
Answer by MathLover1(20850) About Me  (Show Source):
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34.8y+=+x%5E2
comparing to 4py=x%5E2, we have
(h,k)=(0,0)
4p=34.8 => p=34.8%2F4=> p=8.7
then
focus: (0,8.7)
vertex: (0,0)
semi-axis length: 8.7
focal parameter: 2%2A8.7=17.4
eccentricity: 1
directrix:+y+=+-8.7
Focal parameter (the distance between the focus to the directrix) is 17.4.