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well a basic form of the parabola is \[(x - h)^2 = 4a(y - k)\] (h, k) is the vertex and a is the focal length |dw:1438028123648:dw| so the vertex is at the origin so the equation is \[x^2 = 4ay\] substitute a point from the curve (4, 1) into the equation to find the focal length \[4^2 = 4 \times 1 \times a\] ad the the focus is a units about the vertex hope it helps
alright I needed some help, not for someone to write a book lol. 16 = 4 * 1 * a 16 = 4a a = 4 Now what
oh is that the answer?
that is the focal length, the distance from the vertex to the focus.... so if the vertex is (0, 0) the focus is at (0, 0+a)
so its 4?
well the y value in the focus is y = 4 the x value is the line of symmetry x = 0 so the focus is at (0, 4)
I am sorry, so I need the difference between the focus and the vertex, is it 4?
opp... sorrry you are right... I misread the question as where is the focus... the focal length is 4 so the focus is 4 metres above the vertex