Question
Find the focus and directrix for the parabola y=ax^2. What special meaning does the expression 1/(4a) have?
Answers
Damon
I write a parabola that is upright (holds water) as
(x-h)^2 = 4b (y -k)
where my 4 b is your 1/a and h and k are zero so vertex at (0,0)
vertex is at (0,0)
vertex to focus = b = 1/(4a) so focus at ( 0, 1/(4a) )
vertex to directrix = b but below so at y = -1/(4a)
(x-h)^2 = 4b (y -k)
where my 4 b is your 1/a and h and k are zero so vertex at (0,0)
vertex is at (0,0)
vertex to focus = b = 1/(4a) so focus at ( 0, 1/(4a) )
vertex to directrix = b but below so at y = -1/(4a)